Implicit	differentiation	practice	problems	with	answers
  Implicit	differentiation	practice	problems	with	answers	pdf.	
  What	is	implicit	differentiation?	Jenn,	Founder	Calcworkshop®,	15+	Years	Experience	(Licensed	&	Certified	Teacher)	Excellent	question	—	and	that’s	exactly	what	you’re	going	to	learn	in	today’s	calculus	class.	Let’s	go!	Did	you	know	that	implicit	differentiation	is	just	a	method	for	taking	the	derivative	of	a	function	when	x	and	y	are	intermixed?
  Implicit	Vs	Explicit	Functions	But	to	really	understand	this	concept,	we	first	need	to	distinguish	between	explicit	functions	and	implicit	functions.	An	explicit	function	is	an	equation	written	in	terms	of	the	independent	variable,	whereas	an	implicit	function	is	written	in	terms	of	both	dependent	and	independent	variables.	Explicit	Vs.	Implicit	Notice	how
  all	the	explicit	functions	are	solved	for	one	variable	(i.e.,	one	variable	on	the	left-hand-side	and	every	other	term	is	on	the	right)	while	the	implicit	functions	have	the	variables	all	intermixed	on	both	sides	of	the	equation.	This	does	not	leave	an	easy	way	for	us	to	solve	for	one	variable.	How	To	Do	Implicit	Differentiation	In	all	of	our	previous	derivative
  lessons,	we	have	dealt	with	explicit	functions	only,	as	they	are	already	solved	for	one	variable	in	terms	of	another.	But	now	it’s	time	to	learn	how	to	find	the	derivative,	or	rate	of	change,	of	equations	that	contain	one	or	more	variables	and	when	x	and	y	are	intermixed.	Take	the	derivative	of	every	variable.	Whenever	you	take	the	derivative	of	“y”	you
  multiply	by	dy/dx.	Solve	the	resulting	equation	for	dy/dx.	Example	Let’s	use	this	procedure	to	solve	the	implicit	derivative	of	the	following	circle	of	radius	6	centered	at	the	origin.	Implicit	Differentiation	Example	–	Circle	And	that’s	it!	The	trick	to	using	implicit	differentiation	is	remembering	that	every	time	you	take	a	derivative	of	y,	you	must	multiply
  by	dy/dx.	Furthermore,	you’ll	often	find	this	method	is	much	easier	than	having	to	rearrange	an	equation	into	explicit	form	if	it’s	even	possible.	Example	Let’s	look	a	harder	problem	with	trig	where	x’s	and	y’s	are	intermixed.	Implicit	Derivative	–	Trig	And	Exponential	Functions	Example	And	sometimes,	we	will	experience	implicit	functions	with	more
  than	one	y-variable.	All	this	means	is	that	we	will	have	multiple	dy/dx	terms	that	we	will	need	to	collect	in	order	to	simplify,	as	the	following	example	nicely	highlights.	Differentiate	Implicitly	–	Product	Rule	Not	so	bad,	right?!	Implicit	Differentiation	Worksheets	(PDF)	Let’s	put	that	pencil	to	paper	and	try	it	on	your	own.	Video	Tutorial	w/	Full	Lesson	&
  Detailed	Examples	(Video)	Together,	we	will	walk	through	countless	examples	and	quickly	discover	how	implicit	differentiation	is	one	of	the	most	useful	and	vital	differentiation	techniques	in	all	of	calculus.	Get	access	to	all	the	courses	and	over	450	HD	videos	with	your	subscription	Monthly	and	Yearly	Plans	Available	Get	My	Subscription	Now	This
  section	covers:	Introduction	to	Implicit	Differentiation	Up	to	now,	we’ve	differentiated	in	explicit	form,	since,	for	example,	\(y\)	has	been	explicitly	written	as	a	function	of	\(x\).	But	sometimes,	we	can’t	get	an	equation	with	a	“\(y\)”	only	on	one	side;	we	may	have	multiply	“\(y\)’s”	in	the	equation.	In	these	cases,	we	have	to	differentiate	“implicitly”,
  meaning	that	some	“\(y\)’s”	are	“inside”	the	equation.	This	is	called	implicit	differentiation,	and	we	actually	have	to	use	the	chain	rule	to	do	this.	Here’s	an	example	of	an	equation	that	we’d	have	to	differentiate	implicitly:	\(y=7{{x}^{2}}y-2{{y}^{2}}-\sqrt{{xy}}\).		Do	you	see	how	it’d	be	really	difficult	to	get	\(y\)	alone	on	one	side?	The	main	thing
  to	remember	is	when	you	are	differentiating	with	respect	to	“\(x\)”	and	what	you	are	differentiating	only	has	“\(\boldsymbol	{x}\)’s”	in	it	(or	constants),	you	just	get	the	derivative	the	normal	way	(since	the	\(dx\)’s	cancel	out).	But	if	you	are	differentiating	with	respect	to	\(x\),	and	what	you	are	differentiating	has	another	variable	in	it,	like	“\(y\)”,	you
  have	to	multiply	by	\(\displaystyle	\frac{{dy}}{{dx}}\text{	(}{y}’)\):					\(\require{cancel}	\displaystyle	\frac{d}{{dx}}\left(	{{{x}^{2}}}	\right)=2x\cdot	\frac{{\cancel{{dx}}}}{{\cancel{{dx}}}}=2x\):			variables	agree,	so	just	use	the	power	rule	(chain	rule	has	no	effect)	\(\displaystyle	\frac{d}{{dx}}\left(	{{{y}^{2}}}	\right)=2y\cdot
  \frac{{dy}}{{dx}}\):			variables	disagree,	so	use	the	power	rule,	and	then	the	chain	rule	After	we	do	the	differentiation,	we	want	to	solve	for	the		\(\displaystyle	\frac{{dy}}{{dx}}\)	by	getting	it	to	one	side	by	itself	(and	we	may	have	both	\(x\)’s	and	\(y\)’s	on	the	other	side,	which	is	fine).	Here	are	the	steps	for	differentiating	implicitly:	Implicit
  Differentiation	Differentiate	both	sides	of	equation	with	respect	to	\(\boldsymbol{x}\).	When	are	you	differentiating	variables	other	than	\(x\)	(such	as	“\(y\)”),	remember	to	multiply	that	term	by	\(\displaystyle	\frac{{dy}}{{dx}}\text{	(}{y}’)\)!	Move	to	the	left	side	of	the	equation	and	move	all	other	terms	to	the	right	side	(even	if	they	have	\(x\)’s	and	\
  (y\)’s	in	them).	Factor	out	the	\({y}’\)	on	the	left	side	of	the	equation,	and	solve	for	\({y}’\).	Example:					\(\displaystyle	\begin{align}{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}\)	I	know	it	looks	a	bit	scary,	but	it’s	really	not	that	bad!	Let’s	do	some	non-trig	problems	first.	Do	you	see	how	we	have	to	use
  the	chain	rule	a	lot	more?	And	note	how	the	algebra	can	get	really	complicated!	Implicit	Differentiation	Problem	Solution											Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by	implicit	differentiation:	\({{x}^{2}}+{{y}^{2}}=25\)	\(\displaystyle	\begin{align}{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}\)
  Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by	implicit	differentiation:	\(\displaystyle	\frac{{x+y}}{{{{y}^{2}}}}=xy\)	\(\displaystyle	\begin{align}\frac{{x+y}}{{{{y}^{2}}}}&=xy\\\frac{{{{y}^{2}}\left(	{1+{y}’}	\right)-\left(	{x+y}	\right)\left(	{2y{y}’}	\right)}}{{{{y}^{4}}}}&=x{y}’+y\\{{y}^{2}}+{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}
  {y}’&=\left(	{x{y}’+y}	\right){{y}^{4}}\\{{y}^{2}}+{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’&=x{{y}^{4}}{y}’+{{y}^{5}}\\{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’-x{{y}^{4}}{y}’&={{y}^{5}}-{{y}^{2}}\\{y}’\left(	{{{y}^{2}}-2xy-2{{y}^{2}}-x{{y}^{4}}}	\right)&={{y}^{5}}-{{y}^{2}}\\{y}’&=\frac{{{{y}^{5}}-{{y}^{2}}}}
  {{-{{y}^{2}}-2xy-x{{y}^{4}}}}=\frac{{{{y}^{4}}-y}}{{-y-2x-x{{y}^{3}}}}\end{align}\)	Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by	implicit	differentiation:	\(\sqrt{{xy}}+2x{{y}^{2}}=3\)	\(\displaystyle	\begin{align}{{\left(	{xy}	\right)}^{{\frac{1}{2}}}}+2x{{y}^{2}}&=3\\\frac{1}{2}{{\left(	{xy}	\right)}^{{-\frac{1}{2}}}}\left(
  {x{y}’+y}	\right)+2\left(	{x\cdot	2y{y}’+{{y}^{2}}}	\right)=0\\\cancel{{{{{\left(	{xy}	\right)}}^{{\frac{1}{2}}}}}}\left[	{\frac{1}{2}\cancel{{{{{\left(	{xy}	\right)}}^{{-\frac{1}{2}}}}}}\left(	{x{y}’+y}	\right)+2\left(	{2xy{y}’+{{y}^{2}}}	\right)}	\right]&=0\cdot	{{\left(	{xy}	\right)}^{{\frac{1}{2}}}}\,\,\,\,\text{multiply	both	sides	by	}
  {{\left(	{xy}	\right)}^{{\frac{1}{2}}}}\\\frac{1}{2}\left(	{x{y}’+y}	\right)+4xy{y}'{{\left(	{xy}	\right)}^{{\frac{1}{2}}}}+2{{y}^{2}}{{\left(	{xy}	\right)}^{{\frac{1}{2}}}}&=0\\x{y}’+y+8xy{y}'{{\left(	{xy}	\right)}^{{\frac{1}{2}}}}+4{{y}^{2}}{{\left(	{xy}	\right)}^{{\frac{1}{2}}}}&=0\,\,\,\,\,\text{multiply	both	sides	by	}2\\
  {y}’\left(	{8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x}	\right)&=-y-4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}\\{y}’&=\frac{{-y-4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}}}{{8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x}}=-\frac{{y+4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}}}{{8{{x}^{{\frac{3}
  {2}}}}{{y}^{{\frac{3}{2}}}}+x}}\end{align}\)	Let’s	try	some	problems	with	trig	now.	Note	that	with	trig	functions,	it’s	hard	to	know	where	to	stop	simplifying,	so	there	are	several	“correct”	answers.	Implicit	Differentiation	Problem	Solution											Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by	implicit	differentiation:	\(\sin	x+4\cos	\left(	{3y}
  \right)=1\)	\(\displaystyle	\begin{align}\sin	x+4\cos	\left(	{3y}	\right)&=1\\\cos	x+4\cdot	-\sin	\left(	{3y}	\right)\cdot	3\cdot	{y}’&=0\\-12{y}’\sin	\left(	{3y}	\right)&=-\cos	x\\{y}’&=\frac{{-\cos	x}}{{-12\sin	\left(	{3y}	\right)}}=\frac{{\cos	x}}{{12\sin	\left(	{3y}	\right)}}=\frac{1}{{12}}\cos	x\csc	\left(	{3y}	\right)\end{align}\)	Find	\(\displaystyle
  \frac{{dx}}{{dy}}\)	by	implicit	differentiation:	\(\cos	x=x\left(	{2+\csc	y}	\right)\)	\(\displaystyle	\begin{align}\cos	x&=x\left(	{2+\csc	y}	\right)\\-\sin	x&=x\left(	{0+-\csc	y\cot	y\cdot	{y}’}	\right)+\left(	{2+\csc	y}	\right)\left(	1	\right)\\-\sin	x&=-x{y}’\csc	y\cot	y+2+\csc	y\\x{y}’\csc	y\cot	y&=2+\csc	y+\sin	x\\{y}’&=\frac{{2+\csc	y+\sin	x}}{{x\csc
  y\cot	y}}\\&=\frac{2}{{x\csc	y\cot	y}}+\frac{{\cancel{{\csc	y}}}}{{x\cancel{{\csc	y}}\cot	y}}+\frac{{\sin	x}}{{x\csc	y\cot	y}}\,\,\,\,\,(\text{expand)}\\&=\frac{2}{{x\frac{1}{{\sin	y}}\cdot	\frac{{\cos	y}}{{\sin	y}}}}+\frac{1}{{x\cdot	\frac{{\cos	y}}{{\sin	y}}}}+\frac{{\sin	x}}{{x\cdot	\frac{1}{{\sin	y}}\cdot	\frac{{\cos	y}}{{\sin
  y}}}}\,\,\,\,\,\,\text{(put	in	terms	of	sin,	cos)}\\&=\frac{{2{{{\sin	}}^{2}}y}}{{x\cos	y}}+\frac{{\tan	y}}{x}+\frac{{\sin	x\cdot	{{{\sin	}}^{2}}y}}{{x\cos	y}}\\&=\frac{{2\sin	y\tan	y+\tan	y+\sin	x\sin	y\tan	y}}{x}\end{align}\)	(see	how	we	can	get	carried	away	with	simplifying?)	Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by	implicit	differentiation:
  \(\displaystyle	\begin{array}{c}{{\left(	{\sec	\left(	{\pi	x}	\right)+\tan	\left(	{\pi	x}	\right)}	\right)}^{2}}\\=4\end{array}\)	\(\displaystyle	\begin{align}{{\left(	{\sec	\left(	{\pi	x}	\right)+\tan	\left(	{\pi	y}	\right)}	\right)}^{2}}&=4\\2{{\left(	{\sec	\left(	{\pi	x}	\right)+\tan	\left(	{\pi	y}	\right)}	\right)}^{1}}\left(	{\sec	\left(	{\pi	x}	\right)\tan	\left(	{\pi
  x}	\right)\cdot	\pi	+{{{\sec	}}^{2}}\left(	{\pi	y}	\right)\cdot	\pi	{y}’}	\right)&=0\\\frac{{\cancel{{2\left(	{\sec	\left(	{\pi	x}	\right)+\tan	\left(	{\pi	y}	\right)}	\right)}}\left(	{\sec	\left(	{\pi	x}	\right)\tan	\left(	{\pi	x}	\right)\cdot	\pi	+{{{\sec	}}^{2}}\left(	{\pi	y}	\right)\cdot	\pi	{y}’}	\right)}}{{\cancel{{2\left(	{\sec	\left(	{\pi	x}	\right)+\tan	\left(	{\pi
  y}	\right)}	\right)}}}}&=\frac{0}{{2\left(	{\sec	\left(	{\pi	x}	\right)+\tan	\left(	{\pi	y}	\right)}	\right)}}\\\sec	\left(	{\pi	x}	\right)\tan	\left(	{\pi	x}	\right)\cdot	\pi	+{{\sec	}^{2}}\left(	{\pi	y}	\right)\cdot	\pi	{y}’&=0\\{{\sec	}^{2}}\left(	{\pi	y}	\right)\cdot	\pi	{y}’&=-\sec	\left(	{\pi	x}	\right)\tan	\left(	{\pi	x}	\right)\cdot	\pi	\\{y}’&=\frac{{-\sec	\left(
  {\pi	x}	\right)\tan	\left(	{\pi	x}	\right)\cdot	\cancel{\pi	}}}{{{{{\sec	}}^{2}}\left(	{\pi	y}	\right)y\cdot	\cancel{\pi	}}}\\=-\sec	\left(	{\pi	x}	\right)\tan	&\left(	{\pi	x}	\right){{\cos	}^{2}}\left(	{\pi	y}	\right)\end{align}\)	Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by	implicit	differentiation:	\(\displaystyle	\cos	(xy)=5x\)	\(\displaystyle	\begin{align}\cos
  \left(	{xy}	\right)&=5x\\-\sin	\left(	{xy}	\right)\left(	{x{y}’+y}	\right)&=5\\\sin	\left(	{xy}	\right)\left(	{x{y}’+y}	\right)&=-5\\x{y}’\sin	\left(	{xy}	\right)+y\sin	\left(	{xy}	\right)=-5\\x{y}’\sin	\left(	{xy}	\right)&=-5-y\sin	\left(	{xy}	\right)\\{y}’&=-\frac{{5+y\sin	\left(	{xy}	\right)}}{{x\sin	\left(	{xy}	\right)}}=\frac{{-5\csc	\left(	{xy}	\right)+y}}
  {x}\end{align}\)	Equation	of	the	Tangent	Line	with	Implicit	Differentiation	Here	are	some	problems	where	you	have	to	use	implicit	differentiation	to	find	the	derivative	at	a	certain	point,	and	the	slope	of	the	tangent	line	to	the	graph	at	a	certain	point.	The	last	problem	asks	to	find	the	equation	of	the	tangent	line	and	normal	line	(the	line	perpendicular
  to	the	tangent	line	–	take	the	negative	reciprocal	of	the	slope)	at	a	certain	point.	Note	that	we	learned	about	finding	the	Equation	of	the	Tangent	Line	in	the	Equation	of	the	Tangent	Line,	Tangent	Line	Approximation,	and	Rates	of	Change	section.	Implicit	Differentiation	Tangent	Problem	Solution											Find	\(\displaystyle	\frac{{dx}}{{dy}}\)	by
  implicit	differentiation	and	evaluate	the	derivative	at	point	\(\left(	{-2,4}	\right)\):	\({{x}^{2}}+{{y}^{3}}=68\)										\(\displaystyle	\begin{align}{{x}^{2}}+{{y}^{3}}&=68\\2x+3{{y}^{2}}{y}’&=0\\3{{y}^{2}}{y}’&=-2x\\{y}’&=-\frac{{2x}}{{3{{y}^{2}}}}\end{align}\)							\(\displaystyle	\text{At	}\left(	{-2,4}	\right),\,\,\,{y}’=-
  \frac{{2\left(	{-2}	\right)}}{{3{{{\left(	4	\right)}}^{2}}}}=-\frac{{-4}}{{48}}=\frac{1}{{12}}\)	Find	the	slope	of	the	tangent	line	to	the	graph	at	point	\(\left(	{2,1}	\right)\):	\(\displaystyle	{{x}^{3}}+{{y}^{3}}=\frac{9}{2}xy\)	\(\displaystyle	\begin{align}{{x}^{3}}+{{y}^{3}}&=\frac{9}{2}xy\\3{{x}^{2}}+3{{y}^{2}}{y}’&=\frac{9}
  {2}\left(	{x{y}’+y\cdot	1}	\right)\\6{{x}^{2}}+6{{y}^{2}}{y}’&=9\left(	{x{y}’+y}	\right)\\6{{x}^{2}}+6{{y}^{2}}{y}’&=9x{y}’+9y\\6{{y}^{2}}{y}’-9x{y}’&=-6{{x}^{2}}+9y\\{y}’\left(	{6{{y}^{2}}-9x}	\right)&=-6{{x}^{2}}+9y\\{y}’&=\frac{{-6{{x}^{2}}+9y}}{{6{{y}^{2}}-9x}}\\{y}’&=\frac{{-2{{x}^{2}}+3y}}
  {{2{{y}^{2}}-3x}}\end{align}\)		\(\displaystyle	\text{At	}\left(	{2,1}	\right),\,\,\,{y}’=\frac{{-2{{{\left(	2	\right)}}^{2}}+3\left(	1	\right)}}{{2{{{\left(	1	\right)}}^{2}}-3\left(	2	\right)}}=\frac{5}{4}\)	Find	the	equation	of	for	the	tangent	line	and	the	normal	line	to	the	circle	at	the	point	\(\left(	{3,4}	\right)\):	\({{x}^{2}}+{{y}^{2}}=25\)	\
  (\displaystyle	\begin{align}\\{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}\)	\(\displaystyle	\begin{array}{c}\text{Tangent	Line:}\\\text{At	}\left(	{3,4}	\right),\,\,\,{y}’=-\frac{3}{4}\\y-{{y}_{1}}=-\frac{3}{4}\left(	{x-{{x}_{1}}}	\right)\\y-4=-\frac{3}{4}\left(	{x-3}	\right)\\y=-\frac{3}
  {4}x+\frac{{25}}{4}\end{array}\)													\(\displaystyle	\begin{array}{c}\text{Normal	Line:}\\\text{At	}\left(	{3,4}	\right),\,\,\,{y}’=-\frac{3}{4}\\y-{{y}_{1}}=\frac{4}{3}\left(	{x-{{x}_{1}}}	\right)\\y-4=\frac{4}{3}\left(	{x-3}	\right)\\y=\frac{4}{3}x\end{array}\)	Here’s	a	problem	where	we	have	to	use	implicit	differentiation	to	find	the	points
  at	which	the	graph	of	the	equation	has	a	horizontal	and	vertical	tangent	line.	Remember	that	for	the	horizontal	tangent	line,	we	set	the	numerator	to	0,	and	for	the	vertical	tangent	line,	we	set	the	denominator	to	0.	Implicit	Differentiation	and	Horizontal/Vertical	Tangent	Lines												Use	implicit	differentiation	to	find	the	points	at	which	the	graph	of
  the	equation	has	a	horizontal	and	vertical	tangent	line:	\(2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\)		Use	Implicit	Differentiation	to	get	\({y}’\):	\(\displaystyle	\begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\4x+2y{y}’-4+2{y}’+0=0\\2y{y}’+2{y}’=-4x+4\\{y}’\left(	{2y+2}	\right)=-4x+4\end{array}\)	\(\displaystyle	{y}’=\frac{{-4x+4}}
  {{2y+2}}\)	\(\displaystyle	{y}’=\frac{{2-2x}}{{y+1}}\)	Points	at	Horizontal	Tangent	(set	numerator	to	0):	\(\displaystyle	\begin{array}{c}2-2x=0\\x=1\end{array}\)	Now	find	\(y\)	(use	original):	\(\displaystyle	\begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{\left(	1	\right)}^{2}}+{{y}^{2}}-4\left(	1	\right)+2y+2=0\\
  {{y}^{2}}+2y=0\\y\left(	{y+2}	\right)=0\\y=0,\,\,-2\\\left(	{1,0}	\right)\,\,\text{and}\,\,\left(	{1,-2}	\right)\end{array}\)	Points	at	Vertical	Tangent	(set	denominator	to	0):	\(\displaystyle	\begin{array}{c}y+1=0\\y=-1\end{array}\)	Now	find	\(x\)	(use	original):	\(\displaystyle	\begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{x}^{2}}+
  {{\left(	{-1}	\right)}^{2}}-4x+2\left(	{-1}	\right)+2=0\\2{{x}^{2}}-4x+1=0\end{array}\)	\(\displaystyle	x=\frac{{-b\pm	\sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\)	\(\displaystyle	=\frac{{-\left(	{-4}	\right)\pm	\sqrt{{{{{\left(	{-4}	\right)}}^{2}}-4\cdot	2\cdot	1}}}}{{2\cdot	2}}\)	\(\displaystyle	x=\frac{{4\pm	\sqrt{8}}}{4}=\frac{{4\pm	2\sqrt{2}}}
  {4}=\frac{{2\pm	\sqrt{2}}}{2}\)	\(\displaystyle	\left(	{\frac{{2-\sqrt{2}}}{2},\,\,-1}	\right)\,\,\text{and}\,\,\left(	{\frac{{2+\sqrt{2}}}{2},\,\,-1}	\right)\)	Using	Implicit	Differentiation	to	Find	Higher	Order	Derivatives	Here’s	a	problem	where	we	have	to	use	implicit	differentiation	twice	to	find	the	second	derivative	\(\displaystyle
  \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}\,\,\,or\,\,\,{y}”\).		This	one	is	really	tricky,	since	we	need	to	substitute	what	we	got	for	\(\displaystyle	\frac{{dy}}{{dx}}\,\,or\,\,{y}’\)	to	simplify.	Note	that	in	this	example,	we	also	substituted	the	original	function	back	in	to	simplify	further.	(The	reason	I	substituted	at	the	end	is	because	the	resulting	answer
  was	one	of	the	choices	on	a	multiple	choice	test	question.	Your	answer	could	be	in	a	number	of	different	forms.)	Problem	Solution											Use	implicit	differentiation	to	find	\(\displaystyle	\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}\):	\({{y}^{5}}={{x}^{{10}}}\)	(Note	that	this	is	the	same	as	\(\displaystyle	\frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1\))	\
  (\displaystyle	\begin{align}{{y}^{5}}&={{x}^{{10}}}\\5{{y}^{4}}{y}’&=10{{x}^{9}}\\{y}’&=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\end{align}\)								\(\displaystyle	{y}’=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\times	\frac{{xy}}{{xy}}\,=\frac{{10{{x}^{{10}}}y}}{{5x{{y}^{5}}}}=\frac{{10y}}{{5x}}\times
  \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=\frac{{10y}}{{5x}}\times	1\,=\frac{{2y}}{x}\)	Notice	the	cool	trick	to	simplify	before	taking	the	second	derivative.	Since	we	ended	up	with	\(\displaystyle	\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\)	after	differentiating	once,	and	we	know	from	the	original	equation	that	\(\displaystyle
  \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1\),	we	can	multiply	by	\(\displaystyle	\frac{{xy}}{{xy}}\)	to	substitute	and	simplify!	Now	take	the	second	derivative	using	implicit	differentiation	and	the	quotient	rule:	\(\require{cancel}	\displaystyle	\begin{align}{y}’&=\frac{{2y}}{x}\\{y}^{\prime	\prime}&=\frac{{x\cdot	2{y}’-2y\left(	1	\right)}}
  {{{{x}^{2}}}}\\&=\frac{{\cancel{x}\cdot	2\left(	{\frac{{2y}}{{\cancel{x}}}}	\right)-2y}}{{{{x}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,(\text{plug}\,\text{	}{y}’\,\,\text{back	in)}\\&=\frac{{4y-2y}}{{{{x}^{2}}}}=\frac{{2y}}{{{{x}^{2}}}}\end{align}\)	Related	Rates	I	used	to	have	such	a	problem	with	related	rates	problems,	until	I	began	writing	down
  the	steps	to	do	them.	And	it	helps	to	remember	that	the	rates	in	these	problems	typically	are	differentiated	with	respect	to	time,	or	\(\displaystyle	\frac{{d\left(	{\text{something}}	\right)}}{{dt}}\).	Hope	these	steps	help:	Draw	a	picture	and	label	any	amounts	that	could	be	changing.	These	amounts	will	be	variables,	even	if	the	problem	gives	you	an
  amount	at	a	certain	time	for	these	parts	(the	“when”).	In	other	words,	understand	what	is	changing	and	what	isn’t.	I	find	it	helpful	to	draw	an	arrow	to	show	where	the	picture	is	“moving”.	On	the	picture	you’ve	drawn,	label	places	that	are	changing	(you’ll	need	variables),	and	places	that	aren’t	changing	(constants).	For	constants,	you	can	put
  numbers	on	the	picture,	if	they	are	given.	Write	down	exactly	what	the	problem	gives	you,	and	what	you	need	to	solve	for.	For	example,	you	may	write	down	“Find		\(\displaystyle	\frac{{dA}}{{dt}}\)		when	r	=	6”.		Remember	again	that	the	rates	(things	that	are	changing)	have	“dt”	(with	respect	to	time)	in	them	–	pretty	much	always.	Also,	determine
  what	snapshot	in	time	(or	in	another	variable)	the	question	is	asking	for	(the	“when”).	For	the	rates,	remember	that	rates	with	words	like	“filling	up”	means	a	positive	volume	rate,	“emptying	out”	means	a	negative	volume	rate.	Now	we	have	to	figure	out	a	way	to	relate	all	the	variables	together.	Write	an	equation	that	relates	all	of	the	given
  information	and	variables	that	you’ve	written	down.	A	lot	of	times	this	is	a	geometry	equation	for	volume,	area,	or	perimeter.	Also,	you’ll	see	the	Pythagorean	Equation	or	the	trigonometric	functions,	such	as	SOH-CAH-TOA.	Remember	that	shapes	stay	in	proportion,	so	we	may	have	to	use	geometrically	similar	figures	and	set	up	proportions.	Other
  constants	can	be	solved	for	any	instant	of	time;	for	example,	if	you	have	“when	length	is	40“,	you	can	get	the	width	at	that	moment	too	if	you	can,	to	use	as	a	constant.	Simplify	by	trying	to	put	everything	in	as	few	variables	as	possible	before	differentiating	(you	may	have	to	substitute	some	variables	by	solving	in	terms	of	other	variables).	For	example,
  you	may	need	to	relate	length	and	width	together	if	you	have	a	perimeter.	I	find	this	the	hardest	part	–	how	to	know	what	to	plug	in	before	I	differentiate,	and	what	to	plug	in	after.	Other	constants	can	be	solved	for	any	instant	of	time;	for	example,	if	you	have	“when	length	is	40”,	you	can	get	the	width	at	that	moment	too,	to	use	as	a	constant.
  Remember	that	every	variable	you	have	before	differentiating	will	eventually	have	a	“dt”	denominator	after	differentiating,	so	get	rid	of	any	variables	where	rates	of	that	variable	aren’t	given	or	needed	in	the	problem!	The	rule	of	thumb	is	that	when	you	have	values	that	say	“when”	something	happens,	these	are	put	in	after	differentiating.	Use	implicit
  differentiation	to	differentiate	with	respect	to	time.	Hopefully	the	rates		(\(\displaystyle	\frac{{d\left(	{\text{something}}	\right)}}{{dt}}\))		you	have	left	at	this	point	are	rates	that	either	you	have	values	for,	or	need	values	for.	Fill	in	the	equation	with	what	you	know	(for	example,	the	“when”s).	Solve	the	equation	for	what	you	need	(what	you	don’t
  know).	Make	sure	you	are	answering	the	question	that	is	being	asked,	and	the	units	are	correct.	You’d	hate	to	do	all	this	work,	and	then	answer	the	wrong	question.	Remember	when	dealing	with	angles,	make	sure	the	calculator	is	in	the	correct	mode	(radians	or	degrees)!	Here	are	some	examples:	Related	Rates	Problem	Steps	and	Solution											A
  snowball	is	melting	at	a	rate	of	1	cubic	inch	per	minute.	At	what	rate	is	the	radius	changing	when	the	snowball	has	a	radius	of	2	inches?	Draw	a	picture.	Note	that	since	volume	and	radius	are	changing,	we	have	to	use	variables	for	them.	Let	\(r=\)	radius,	\(V=\)	volume.	Define	what	we	have	and	what	we	want	when.	We	have:	\(\displaystyle
  \frac{{dV}}{{dt}}=-1\)	(since	volume	is	decreasing).	We	want:	find	\(\displaystyle	\frac{{dr}}{{dt}}\)	when	\(r=2\).	Determine	what	equation	we	need:	we	need	to	relate	the	radius	of	a	circle	to	its	volume	(sphere):	\(\displaystyle	{{V}_{{sphere}}}=\frac{4}{3}\pi	{{r}^{3}}\).	Make	sure	that	all	variables	will	differentiate	to	a	rate	we	either	have
  or	need:	yes!	Differentiate	with	respect	to	time:	\(\displaystyle	\require{cancel}	\frac{{dV}}{{dt}}=\cancel{3}\cdot	\frac{4}{{\cancel{3}}}\pi	\,{{r}^{2}}\frac{{dr}}{{dt}}=4\pi	\,{{r}^{2}}\frac{{dr}}{{dt}}\)	.	Plug	in	what	we	know	at	the	time	we	know	it,	and	solve	for	what	we	need:	\(\displaystyle	4\pi	{{r}^{2}}\frac{{dr}}
  {{dt}};\,\,\,\,\,-1=4\pi	{{\left(	2	\right)}^{2}}\cdot	\frac{{dr}}{{dt}};\,\,\,\,\,\frac{{dr}}{{dt}}=\frac{{-1}}{{16\pi	}}\,\,\approx	\,\,-.0199\,\,\text{in/min}\).	Make	sure	we’re	answering	the	question:		yes,	–.0199	inches	per	minute	is	the	rate	the	radius	is	changing.	It	makes	sense	that	it	is	negative,	since	radius	is	decreasing	along	with	the	volume.	A
  40-foot	ladder	is	resting	against	a	wall,	and	its	base	is	slipping	on	the	floor	(away	from	the	wall)	at	a	rate	of	2	feet	per	minute.	Find	the	rate	of	change	of	the	height	of	the	ladder	at	the	time	when	the	base	is	20	feet	from	the	base	of	the	wall.	Draw	a	picture.	Note	that	since	height	(\(y\))	and	distance	from	base	(\(x\))	are	changing,	we	have	to	use
  variables	for	them.	The	ladder	(slanted)	will	be	the	hypotenuse	of	a	right	triangle	formed	by	the	wall,	the	floor,	and	the	ladder.	Define	what	we	have	and	what	we	want	when.	We	have:	\(\displaystyle	\frac{{dx}}{{dt}}=2\).	We	want:	find	\(\displaystyle	\frac{{dy}}{{dt}}\)	when	\(x=20\).	Determine	what	equation	we	need:		we	need	to	relate	the
  variables	together;	let’s	use	the	Pythagorean	Theorem:	\({{x}^{2}}+{{y}^{2}}={{40}^{2}}\).	Make	sure	that	all	variables	will	differentiate	to	a	rate	we	either	have	or	need:	yes!	Differentiate	with	respect	to	time:	\(\displaystyle	2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0\).	Plug	in	what	we	know	at	the	time	we	know	it,	and	solve	for	what	we
  need.	The	tricky	part	here	is	that	we	need	to	also	use	the	Pythagorean	Theorem	again	to	get	what	\(y\)	(height)	will	be	when	\(x=20\)	(distance	from	base	of	wall):		\({{20}^{2}}+{{y}^{2}}={{40}^{2}};\,\,\,\,y=\sqrt{{1200}}\approx	34.641\).	Now	plug	everything	in:					\(\displaystyle	2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0;\,\,\,\,2\left(	{20}
  \right)\left(	2	\right)+2\left(	{34.641}	\right)\frac{{dy}}{{dt}}=0\)									\(\displaystyle	\frac{{dy}}{{dt}}=-\frac{{2\left(	{20}	\right)\left(	2	\right)}}{{2\left(	{34.641}	\right)}}\approx	-1.155\,\,\text{ft/min}\).	Make	sure	we’re	answering	the	question:		yes,	–1.155	feet	per	minute	is	the	rate	the	height	of	the	ladder	is	changing.	It	makes	sense	that	it
  is	negative,	since	the	ladder	is	slipping	down	the	wall.	Here	are	more	problems:	Related	Rates	Problem	Steps	and	Solution											Water	is	being	poured	into	a	cylindrical	can	that	is	20	inches	in	height	and	has	a	radius	of	8	inches.	The	water	is	being	poured	at	a	rate	of	3	cubic	inches	per	second.	How	fast	is	the	height	of	the	water	in	the	can	changing
  when	the	height	is	8	inches	deep?	Draw	a	picture.	Note	that	since	the	radius	and	height	of	the	actual	can	is	not	changing,	we	can	use	constants	for	them.	Since	the	height	(and	volume)	of	the	water	inside	is	changing,	we	have	to	use	variables:	Let	\(h=\)	height	of	water,	\(V=\)	volume	of	water.	Define	what	we	have	and	what	we	want	when.	We	have:
  height	of	can	\(=20\),	radius	of	can	(and	water)	\(=8\),	\(\displaystyle	\frac{{dV}}{{dt}}=3\)	(rate	of	volume	of	water	in	can).	We	want:	find	\(\displaystyle	\frac{{dh}}{{dt}}\)	(rate	of	height	of	water)	when	\(h=8\)	(height	of	water).	Determine	what	equation	we	need:	We	need	to	relate	the	radius	(a	constant)	and	height	of	a	cylinder	to	its	volume:	\
  ({{V}_{{cylinder}}}=\pi	{{r}^{2}}h\).	Differentiate	with	respect	to	time:	\(\displaystyle	\frac{{dV}}{{dt}}=\pi	{{r}^{2}}\frac{{dh}}{{dt}}\).	Plug	in	what	we	know	at	the	time	we	know	it,	and	solve	for	what	we	need:	\(\displaystyle	\frac{{dV}}{{dt}}=\pi	{{r}^{2}}\frac{{dh}}{{dt}};\,\,\,\,\,3=\pi	{{\left(	8	\right)}^{2}}\frac{{dh}}
  {{dt}};\,\,\,\,\frac{{dh}}{{dt}}\approx	.015\,\,\text{ft/sec}\)	Make	sure	we’re	answering	the	question:	Yes,	.015	feet	per	second	is	the	rate	the	height	of	the	water	is	changing.	It	makes	sense	that	it	is	positive,	since	the	can	is	filling	up.	Related	Rates	Problem	Steps	and	Solution											A	woman	5	feet	tall	walks	at	a	rate	of	6	feet	per	second	away	from
  a	lamppost.	When	the	woman	is	6	feet	from	the	lamppost,	her	shadow	is	8	feet	tall.	At	what	rate	is	the	length	of	her	shadow	changing	when	she	is	12	feet	from	the	lamppost?	At	what	rate	is	the	tip	of	her	shadow	changing	when	she	is	12	feet	from	the	lamppost?	Draw	a	picture.	Remember	that	the	woman’s	shadow	is	on	the	ground	in	front	of	her;	we’ll
  use	\(y\)	for	this	distance.	The	distance	from	the	base	of	the	lamppost	to	the	woman	is	\(x\).	We	need	variables	for	both	of	these	distances,	since	they	are	changing.	Define	what	we	have	and	what	we	want	when.	We	have	the	rate	the	woman	walks	from	the	lamppost,	\(\displaystyle	\frac{{dx}}{{dt}}=6\).	Now	this	is	tricky:	to	get	the	rate	of	the	length
  of	her	shadow	increasing,	we	want	,\(\displaystyle	\frac{{dy}}{{dt}}\)	but	if	we	want	the	rate	of	the	tip	of	her	shadow	increasing,	we	want	\(\displaystyle	\frac{{d\left(	{x+y}	\right)}}{{dt}}\).	(This	is	because	we	need	to	measure	the	tip	of	the	shadow	in	reference	to	the	base	of	the	lamppost).	We	also	have	the	height	of	the	woman	(5	feet),	and	we
  know	that	when	the	woman	is	6	feet	from	the	lamppost	\((x=6),\)	and	her	shadow	is	8	feet	(\(y=8\)).	It	turns	out	that	we	will	need	the	height	\(h\)	of	the	lamppost	to	get	the	equation	we	need	to	differentiate,	so	from	similar	triangles,	we	can	get	this:	(see	last	triangle	to	the	left):	\(\displaystyle	\frac{{\text{side	of	large	}\Delta	}}{{\text{side	of	small
  }\Delta	}}=\frac{{\text{bottom	of	large	}\Delta	}}{{\text{bottom	of	small	}\Delta	}}:\,\,\,\,\,\,\frac{h}{5}=\frac{{14}}{8};\,\,\,h=8.75\)	Determine	what	equation	we	need:		We	need	to	relate	\(x\)	and	\(y\)	to	what	we	know	(\(h\)	and	woman’s	height),	which	we	can	do	using	similar	triangles	again:	\(\displaystyle	\frac{{\text{8}\text{.75}}}
  {5}=\frac{{x+y}}{y},\)	or	(simplified),	\(3.75y=5x\).	Make	sure	that	all	variables	will	differentiate	to	a	rate	we	either	have	or	need:	yes!	Differentiate	with	respect	to	time:	\(\displaystyle	3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}}\).	Plug	in	what	we	know	at	the	time	we	know	it,	and	solve	for	what	we	need:	Now	plug	in:	\(\displaystyle
  3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}};\,\,\,\,3.75\frac{{dy}}{{dt}}=5\left(	6	\right);\,\,\,\frac{{dy}}{{dt}}=8\)	feet	per	second.	This	is	the	rate	of	the	length	of	the	shadow.	To	get	the		rate	of	the	tip	of	the	shadow,	we	need	\(\displaystyle	\frac{{d\left(	{x+y}	\right)}}{{dt}}\),	which	is	actually	\(\displaystyle	\frac{{dx}}{{dt}}+\frac{{dy}}
  {{dt}}\)	(note	that	could	have	also	defined	a	variable	to	represent	\(x	+	y\),	and	do	some	subtraction	in	setting	it	up).	The	rate	of	the	tip	of	the	shadow	is	\(6+8=14\)	feet	per	second.	Here	are	a	few	that	involve	right	triangle	trigonometry:	Related	Rates	Problems	with	Trig	Steps	and	Solution											A	hot	air	balloon	is	rising	straight	up	and	is	tracked	by
  a	range	finder	400	feet	from	point	of	lift-off.	When	the	range	finder’s	angle	of	elevation	is	\(\displaystyle	\frac{\pi	}{3}\),	the	angle	is	increasing	at	a	rate	of	.15	radians	per	minute.	How	fast	is	the	balloon	rising	at	that	time?	Draw	a	picture.	Note	that	since	height	and	the	angle	of	elevation	are	changing,	we	have	to	use	variables	for	them.	Let	\
  (h=\)	height,	\(\theta	=\)	angle	of	elevation.	Define	what	we	have	and	what	we	want	when.	We	have:	\(\displaystyle	\frac{{d\theta	}}{{dt}}=.15\)	(in	radians).	We	want:	find	\(\displaystyle	\frac{{dh}}{{dt}}\)	when	\(\displaystyle	\theta	=\frac{\pi	}{3}\).	We	also	know	the	distance	on	the	ground	from	the	range	finder	to	the	balloon	is	400.	Determine
  what	equation	we	need:	We	need	to	relate	the	height	and	base	of	the	right	triangle	to	the	angle	of	elevation;	we	can	use	trigonometry:	\(\displaystyle	\tan	\theta	=\frac{h}{{400}}\).	Make	sure	that	all	variables	will	differentiate	to	a	rate	we	either	have	or	need:	yes!	Differentiate	with	respect	to	time:	\(\displaystyle	\frac{{d\left(	{\tan	\theta	}	\right)}}
  {{dt}}=\frac{{d\left(	{\frac{1}{{400}}}	\right)}}{{dt}};\,\,\,\,\,{{\sec	}^{2}}\theta	\frac{{d\theta	}}{{dt}}=\frac{1}{{400}}\frac{{dh}}{{dt}}\).	Plug	in	what	we	know	at	the	time	we	know	it,	and	solve	for	what	we	need	(calculator	in	radians):	\(\displaystyle	\begin{align}{{\sec	}^{2}}\theta	\frac{{d\theta	}}{{dt}}&=\frac{1}
  {{400}}\frac{{dh}}{{dt}};\,\,\,\,\,\left(	{{{{\sec	}}^{2}}\left(	{\frac{\pi	}{3}}	\right)}	\right)\left(	{.15}	\right)=\frac{1}{{400}}\frac{{dh}}{{dt}}\\{{2}^{2}}\left(	{.15}	\right)&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,\frac{{dh}}{{dt}}=240\,\,\text{rad/min}\end{align}\)			\(\displaystyle	(\text{note	that	}\sec	\left(	{\frac{\pi	}{3}}
  \right)=\frac{1}{{\cos	\left(	{\frac{\pi	}{3}}	\right)}}=\frac{1}{{\frac{1}{2}}}=2)\)	Make	sure	we’re	answering	the	question:	Yes,	240	radians	per	minute	is	the	rate	the	angle	of	elevation	is	changing.	It	makes	sense	that	it	is	positive	since	the	angle	is	increasing	along	with	the	height.	Two	women	are	35	feet	apart.	One	of	them	starts	walking	south
  at	a	rate	so	that	the	angle	below	in	the	diagram	is	changing	at	a	constant	rate	of	0.02	radians	per	minute.	At	what	rate	is	the	distance	between	the	two	women	changing	when	\(\theta	=.5\)	radians?	Draw	a	picture.	Note	that	since	\(\theta	\)	and	the	distance	between	the	women	are	changing,	we	have	to	use	variables	for	them.	Let	\(x=\)	the	distance
  between	the	two	women.	Define	what	we	have	and	what	we	want	when.	We	have:	\(\displaystyle	\frac{{d\theta	}}{{dt}}=\,\,.02\)	(in	radians).	We	want:	find	\(\displaystyle	\frac{{dx}}{{dt}}\)	when	\(\displaystyle	\theta	=.5\).	We	also	know	the	original	distance	between	the	two	women	is	35.	Determine	what	equation	we	need:	We	need	to	relate	the
  given	angle	to	the	sides	of	the	triangle;	we	can	use	trigonometry:		\(\displaystyle	\cos	\theta	=\frac{{35}}{x}\).	As	we’ll	see	below,	we’ll	also	need	to	get	\(x\)	and	relate	it	back	to	\(\theta	\)	when	\(\theta	=.5\);		we’ll	use	this	same	equation:	\(\displaystyle	\cos	\left(	{.5}	\right)=\frac{{35}}{x};\,\,x=39.882\).	Differentiate	with	respect	to	time:	\
  (\displaystyle	\frac{{d\left(	{\cos	\theta	}	\right)}}{{dt}}=\frac{{d\left(	{35{{x}^{{-1}}}}	\right)}}{{dt}};\,\,\,\,\,-\sin	\theta	\frac{{d\theta	}}{{dt}}=-35{{x}^{{-2}}}\frac{{dx}}{{dt}}\).	Plug	in	what	we	know	at	the	time	we	know	it,	and	solve	for	what	we	need	(calculator	in	radians):	\(\displaystyle	\begin{align}\,-\sin	\theta	\frac{{d\theta	}}
  {{dt}}&=-35{{x}^{{-2}}}\frac{{dx}}{{dt}};\,\,\,\,\,\left(	{-\sin	\left(	{.5}	\right)}	\right)\left(	{.02}	\right)=-35{{\left(	{39.882}	\right)}^{{-2}}}\frac{{dx}}{{dt}}\,\\\frac{{dx}}{{dt}}&=\frac{{\left(	{-\sin	\left(	{.5}	\right)}	\right)\left(	{.02}	\right){{{\left(	{39.882}	\right)}}^{2}}}}{{-35}}=.436\,\,\text{ft/min}\end{align}\)	Make	sure	we’re
  answering	the	question:	Yes,	.436	feet	per	minute	is	the	rate	the	distance	between	the	two	women	is	changing.	It	makes	sense	that	it	is	positive	since	the	angle	between	them	is	increasing.	Understand	these	problems,	and	practice,	practice,	practice!	Click	on	Submit	(the	arrow	to	the	right	of	the	problem)	to	solve	this	problem.	You	can	also	type	in
  more	problems,	or	click	on	the	3	dots	in	the	upper	right	hand	corner	to	drill	down	for	example	problems.	If	you	click	on	“Tap	to	view	steps”,	you	will	go	to	the	Mathway	site,	where	you	can	register	for	the	full	version	(steps	included)	of	the	software.		You	can	even	get	math	worksheets.	You	can	also	go	to	the	Mathway	site	here,	where	you	can	register,
  or	just	use	the	software	for	free	without	the	detailed	solutions.		There	is	even	a	Mathway	App	for	your	mobile	device.		Enjoy!	On	to	Curve	Sketching	–	you’re	ready!		
  Implicit	differentiation	is	an	important	concept	to	know	in	calculus.	This	quiz/worksheet	will	help	you	test	your	understanding	of	it	and	let	you	put	your	skills	to	the	test	with	practice	problems.	28/01/2021	·	Implicit	Differentiation.	Implicit	differentiation	is	a	method	that	makes	use	of	the	chain	rule	to	differentiate	implicitly	defined	functions.	It	is
  generally	not	easy	to	find	the	function	explicitly	and	then	differentiate.	Instead,	we	can	totally	differentiate	f(x,	y)	and	then	solve	the	rest	of	the	equation	to	find	the	value	of	.	Even	when	it	...	Implicit	differentiation	is	nothing	more	than	a	special	case	of	the	well-known	chain	rule	for	derivatives.	The	majority	of	differentiation	problems	in	first-year
  calculus	involve	functions	y	written	EXPLICITLY	as	functions	of	x	.	For	…	04/02/2018	·	Section	3-3	:	Differentiation	Formulas.	For	problems	1	–	12	find	the	derivative	of	the	given	function.	f	(x)	=	6x3−9x	+4	f	(	x)	=	6	x	3	−	9	x	+	4	Solution.	y	=	2t4−10t2	+13t	y	=	2	t	4	−	10	t	2	+	13	t	Solution.	g(z)	=	4z7−3z−7	+9z	g	(	z)	=	4	z	7	−	3	z	−	7	+	9	z
  Solution.	h(y)	=	y−4	−9y−3+8y−2	+12	h	(	y)	=	y	−	4	−	9	...	Differential	Equations:	Problems	with	Solutions	By	Prof.	Hernando	Guzman	Jaimes	(University	of	Zulia	-	Maracaibo,	Venezuela)	04/02/2018	·	Section	3-3	:	Differentiation	Formulas.	For	problems	1	–	12	find	the	derivative	of	the	given	function.	f	(x)	=	6x3−9x	+4	f	(	x)	=	6	x	3	−	9	x	+	4
  Solution.	y	=	2t4−10t2	+13t	y	=	2	t	4	−	10	t	2	+	13	t	Solution.	g(z)	=	4z7−3z−7	+9z	g	(	z)	=	4	z	7	−	3	z	−	7	+	9	z	Solution.	h(y)	=	y−4	−9y−3+8y−2	+12	h	(	y)	=	y	−	4	−	9	...	How	implicit	differentiation	can	be	used	the	find	the	derivatives	of	equations	that	are	not	functions,	calculus	lessons,	examples	and	step	by	step	solutions,	What	is	implicit
  differentiation,	Find	the	second	derivative	using	implicit	differentiation	.	Implicit	Differentiation.	In	these	lessons,	we	will	learn	how	implicit	differentiation	can	be	used	the	find	the	derivatives	of	equations	that	…	Differential	Equations:	Problems	with	Solutions	By	Prof.	Hernando	Guzman	Jaimes	(University	of	Zulia	-	Maracaibo,	Venezuela)	How
  implicit	differentiation	can	be	used	the	find	the	derivatives	of	equations	that	are	not	functions,	calculus	lessons,	examples	and	step	by	step	solutions,	What	is	implicit	differentiation,	Find	the	second	derivative	using	implicit	differentiation	…	Differentiation	questions	with	answers	are	provided	here	for	students	of	Class	11	and	Class	12.	Differentiation
  is	an	important	topic	for	11th	and	12th	standard	students	as	these	concepts	are	further	included	in	higher	studies.	The	problems	prepared	here	are	as	per	the	CBSE	board	and	NCERT	curriculum.	Practising	these	questions	will	help	students	to	solve	hard	problems	and	to	…	Differentiation	questions	with	answers	are	provided	here	for	students	of
  Class	11	and	Class	12.	Differentiation	is	an	important	topic	for	11th	and	12th	standard	students	as	these	concepts	are	further	included	in	higher	studies.	The	problems	prepared	here	are	as	per	the	CBSE	board	and	NCERT	curriculum.	Practising	these	questions	will	help	students	to	solve	hard	problems	and	to	…	28/01/2021	·	Implicit	Differentiation.
  Implicit	differentiation	is	a	method	that	makes	use	of	the	chain	rule	to	differentiate	implicitly	defined	functions.	It	is	generally	not	easy	to	find	the	function	explicitly	and	then	differentiate.	Instead,	we	can	totally	differentiate	f(x,	y)	and	then	solve	the	rest	of	the	equation	to	find	the	value	of	.	Even	when	it	...	Implicit	differentiation	is	an	important
  concept	to	know	in	calculus.	This	quiz/worksheet	will	help	you	test	your	understanding	of	it	and	let	you	put	your	skills	to	the	test	with	practice	problems.	Implicit	differentiation	is	nothing	more	than	a	special	case	of	the	well-known	chain	rule	for	derivatives.	The	majority	of	differentiation	problems	in	first-year	calculus	involve	functions	y	written
  EXPLICITLY	as	functions	of	x	.	For	example,	if.	.	However,	some	functions	y	are	written	IMPLICITLY	as	functions	of	x	.