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Implicit Differentiation
Related Topics
Wize University Calculus 1 Textbook > Derivatives
Implicit Differentiation
4 Activities
Evaluate the slope of the tangent line to the curve at the given point.
x
3
y
2
+
x
tan
y
=
4
a
t
P
(
x
,
y
)
\displaystyle x^3y^2+x\tan{y}=4 \, at\, P(x,y)
x
3
y
2
+
x
tan
y
=
4
a
t
P
(
x
,
y
)
−
3
x
2
y
2
−
tan
(
y
)
2
x
3
y
+
x
sec
2
(
y
)
\dfrac{-3x^2y^2-\tan(y)}{2x^3y+ x\sec^2(y)}
2
x
3
y
+
x
sec
2
(
y
)
−
3
x
2
y
2
−
tan
(
y
)
y
′
=
−
3
x
2
y
2
+
tan
(
y
)
2
x
3
y
+
x
sec
2
(
y
)
y' = \dfrac{-3x^2y^2+\tan(y)}{2x^3y+ x\sec^2(y)}
y
′
=
2
x
3
y
+
x
sec
2
(
y
)
−
3
x
2
y
2
+
tan
(
y
)
y
′
=
3
x
2
y
2
+
tan
(
y
)
2
x
3
y
+
x
sec
2
(
y
)
y' = \dfrac{3x^2y^2+\tan(y)}{2x^3y+ x\sec^2(y)}
y
′
=
2
x
3
y
+
x
sec
2
(
y
)
3
x
2
y
2
+
tan
(
y
)
y
′
=
−
3
x
2
y
2
−
tan
(
y
)
2
x
3
y
−
x
sec
2
(
y
)
y' = \dfrac{-3x^2y^2-\tan(y)}{2x^3y- x\sec^2(y)}
y
′
=
2
x
3
y
−
x
sec
2
(
y
)
−
3
x
2
y
2
−
tan
(
y
)
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Check Submission
More Implicit Differentiation Questions:
Implicit differentiation
Differentiate
e
x
y
−
tan
y
=
3
x
3
+
2
x
e^{xy}-\tan y=\frac{3}{x^3}+2x
e
x
y
−
tan
y
=
x
3
3
+
2
x
Implicit Differentiation: Tangent line
Find the equation of the tangent to the graph
3
y
3
+
x
=
x
2
+
1
3y^3+x=x^2+1
3
y
3
+
x
=
x
2
+
1
at
x
=
2
x=2
x
=
2
.
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Implicit Differentiation: Logarithmic Differentiation
Find the equation of the tangent line to
(
sin
(
x
y
)
)
x
=
x
1
/
4
(\sin(xy))^x = x^{1/4}
(
sin
(
x
y
)
)
x
=
x
1/4
at the point
(
1
2
,
π
2
)
\left(\dfrac{1}{2},\dfrac{\pi}{2}\right)
(
2
1
,
2
π
)
.
Implicit Second Derivative
Find
d
2
y
d
x
2
\dfrac{d^2y}{dx^2}
d
x
2
d
2
y
at the point
(
2
,
2
)
(2,2)
(
2
,
2
)
given
x
y
+
y
2
=
2
xy+y^2=2
x
y
+
y
2
=
2
Implicit differentiation
Differentiate
e
x
y
−
tan
y
=
3
x
3
+
2
x
e^{xy}-\tan y=\frac{3}{x^3}+2x
e
x
y
−
tan
y
=
x
3
3
+
2
x
Implicit Differentiation
Find the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point (1,9).
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Implicit Differentiation
Find
y
'' if
x
3
+
y
3
=
4.
x^3+y^3=4.
x
3
+
y
3
=
4.
Implicit Differentiation
Find
d
y
d
x
\displaystyle \frac{\text{d}y}{\text{d}x}
d
x
d
y
for
y
3
=
2
x
y^3=2x
y
3
=
2
x
.
Implicit Differentiation
Find the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point (1,9).
Implicit differentiation
Find the derivative of the following:
y
e
x
+
y
2
=
x
2
ye^{x+y^2}=x^2
y
e
x
+
y
2
=
x
2
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Find the tangent to the curve 4𝑦
2
− 𝑒
𝑥
+ 𝑥 = 4𝑥 + 3 at the point where 𝑥 = 0 and 𝑦 is positive
Implicit Differentiation
Evaluate the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point
P
(
1
,
9
)
P(1,9)
P
(
1
,
9
)
.
Implicit Differentiation: Tangent line
Find the equation of the tangent to the graph
3
y
3
+
x
=
x
2
+
1
3y^3+x=x^2+1
3
y
3
+
x
=
x
2
+
1
at
x
=
2
x=2
x
=
2
.
Implicit Differentiation: Tangent line
Find the equation of the tangent to the graph
3
y
3
+
x
=
x
2
+
1
3y^3+x=x^2+1
3
y
3
+
x
=
x
2
+
1
at
x
=
2
x=2
x
=
2
.
Implicit Differentiation
Evaluate the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point
P
(
1
,
9
)
P(1,9)
P
(
1
,
9
)
.
Implicit Differentiation
For the implicit equation
x
4
+
x
3
y
−
x
y
2
=
1
x^4+x^3y-xy^2=1
x
4
+
x
3
y
−
x
y
2
=
1
, where y =
f
(
x
), find
y
'.
Implicit Differentiation: Tangent line
Find the equation of the tangent to the graph
3
y
3
+
x
=
x
2
+
1
3y^3+x=x^2+1
3
y
3
+
x
=
x
2
+
1
at
x
=
2
x=2
x
=
2
.
Implicit Differentiation
Find all the equations of the tangent lines to the curve
y
=
x
3
+
x
2
−
x
y=x^3+x^2-x
y
=
x
3
+
x
2
−
x
that are parallel to the x-axis.
Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
at
x
=
0
x=0
x
=
0
if
x
cos
(
y
2
)
=
y
cos
x
x\cos\left(\frac{y}{2}\right)=y\cos x
x
cos
(
2
y
)
=
y
cos
x
.
Implicit Differentiation: Tangent line
Find the equation of the tangent to the graph
3
y
3
+
x
=
x
2
+
1
3y^3+x=x^2+1
3
y
3
+
x
=
x
2
+
1
at
x
=
2
x=2
x
=
2
.
Implicit differentiation
Differentiate
e
x
y
−
tan
y
=
3
x
3
+
2
x
e^{xy}-\tan y=\frac{3}{x^3}+2x
e
x
y
−
tan
y
=
x
3
3
+
2
x
Implicit Differentiation
Given
x
2
+
y
=
y
2
−
4
x^2+y=y^2-4
x
2
+
y
=
y
2
−
4
, find
y
′
y'
y
′
.
Implicit Differentiation
Find
d
y
d
x
\displaystyle \frac{\text{d}y}{\text{d}x}
d
x
d
y
for
y
3
=
2
x
y^3=2x
y
3
=
2
x
.
Find the point(s) where
x
2
−
4
x
+
6
y
+
y
2
=
−
9
x^2-4x+6y+y^2 = −9
x
2
−
4
x
+
6
y
+
y
2
=
−
9
has horizontal and vertical tangent lines.
Find
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
for the function
y
cos
x
=
x
2
+
cos
y
y\cos x=x^2+\cos y
y
cos
x
=
x
2
+
cos
y
Logarithmic for Implicitly Defined Function
Find
d
y
/
d
x
dy/dx
d
y
/
d
x
at the point
(
1
,
1
)
(1,1)
(
1
,
1
)
given
x
y
=
y
x
x^y=y^x
x
y
=
y
x
Practice: Tangent and Normal Lines with Implicit
Q:
\textbf{Q:}
Q:
Find the equation of the tangent and the normal lines to the curve
x
2
cos
2
y
−
sin
y
=
0
x^2\cos^2y-\sin y=0
x
2
cos
2
y
−
sin
y
=
0
at the point
(
0
,
π
)
(0,\pi)
(
0
,
π
)
.
Implicit Second Derivative
Find
d
2
y
d
x
2
\dfrac{d^2y}{dx^2}
d
x
2
d
2
y
at the point
(
2
,
2
)
(2,2)
(
2
,
2
)
given
x
y
+
y
2
=
2
xy+y^2=2
x
y
+
y
2
=
2
Implicit Differentiation
Evaluate the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point
P
(
1
,
9
)
P(1,9)
P
(
1
,
9
)
.
Implicit with Trig
Find
d
y
/
d
x
dy/dx
d
y
/
d
x
for
x
y
=
tan
(
x
+
y
2
)
\displaystyle xy=\tan(x+y^{2})
x
y
=
tan
(
x
+
y
2
)
Implicit Differentiation
Let
x
x
x
,
y
y
y
satisfy the equation
x
y
=
6
y
−
x
+
1
x^y = 6y - x + 1
x
y
=
6
y
−
x
+
1
. Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
at the point
(
x
,
y
)
=
(
1
,
1
/
6
)
(x, y) = (1, 1/6)
(
x
,
y
)
=
(
1
,
1/6
)
.
Derivatives: Logarithmic Functions
Compute the derivative of
f
(
x
)
=
x
x
+
1
f(x) = x^{x + 1}
f
(
x
)
=
x
x
+
1
. Remember that
log
x
=
log
e
x
=
ln
x
\log x = \log_e x = \ln x
lo
g
x
=
lo
g
e
x
=
ln
x
.
Tangent lines: Implicit Differentiation
An equation of the tangent line to the curve
x
3
+
x
ln
y
+
y
4
=
3
x
3
2
+
e
x
x^3+x\ln y+y^4=3x^{\frac{3}{2}}+e^x
x
3
+
x
ln
y
+
y
4
=
3
x
2
3
+
e
x
at
x
=
0
x=0
x
=
0
is
Implicit Differentiation: Logarithmic Functions
For the following equations, solve for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
:
a)
y
=
(
5
x
2
+
3
)
sin
(
x
)
y = (5x^2 + 3)^{\sin(x)}
y
=
(
5
x
2
+
3
)
s
i
n
(
x
)
b)
x
2
y
+
3
y
2
x
2
=
x
+
y
x^2y + 3y^2 x^2 = x + y
x
2
y
+
3
y
2
x
2
=
x
+
y
Evaluate the slope of the tangent line to the curve at the given point.
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
\displaystyle \sqrt{xy}=x^3y-6\,\, at \,\,P(1,9)
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
Implicit differentiation
Find the derivative of the following:
y
e
x
+
y
2
=
x
2
ye^{x+y^2}=x^2
y
e
x
+
y
2
=
x
2
Implicit differentiation
Find the derivative of the following:
sin
(
x
+
y
)
−
sec
(
x
2
+
y
2
)
=
y
\sin(x+y)-\mathrm{\sec}(x^2+y^2)=y
sin
(
x
+
y
)
−
sec
(
x
2
+
y
2
)
=
y
Implicit Differentiation
Evaluate the slope of the tangent line to the curve at the given point.
x
3
+
y
3
=
y
sin
x
+
1
,
at
P
(
0
,
1
)
x^3+y^3=y\sin{x}+1,\,\, \text{at}\,\,P(0,1)
x
3
+
y
3
=
y
sin
x
+
1
,
at
P
(
0
,
1
)
Evaluate the slope of the tangent line to the curve at the given point.
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
\displaystyle \sqrt{xy}=x^3y-6\,\, at \,\,P(1,9)
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
Evaluate the slope of the tangent line to the curve at the given point.
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
\displaystyle \sqrt{xy}=x^3y-6\,\, at \,\,P(1,9)
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
Find the equation of the tangent line to the equation
x
3
+
ln
y
=
8
x^3+\ln y=8
x
3
+
ln
y
=
8
at the point (2, 1)
Evaluate the slope of the tangent line to the curve at the given point.
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
\displaystyle \sqrt{xy}=x^3y-6\,\, at \,\,P(1,9)
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
Implicit Differentiation: Logarithmic Differentiation
Find the equation of the tangent line to
(
sin
(
x
y
)
)
x
=
x
1
/
4
(\sin(xy))^x = x^{1/4}
(
sin
(
x
y
)
)
x
=
x
1/4
at the point
(
1
2
,
π
2
)
\left(\dfrac{1}{2},\dfrac{\pi}{2}\right)
(
2
1
,
2
π
)
.
Implicit Differentiation
Find the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point (1,9).
Implicit Differentiation
Find
y
'' if
x
3
+
y
3
=
4.
x^3+y^3=4.
x
3
+
y
3
=
4.
Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
given that
e
x
y
−
x
y
=
2
x
2
+
1
e^{xy}-xy=2x^2+1
e
x
y
−
x
y
=
2
x
2
+
1
final114
Find
y
′
y'
y
′
for the function
x
4
+
x
3
y
−
x
y
2
=
1
x^4+x^3y-xy^2=1
x
4
+
x
3
y
−
x
y
2
=
1
Differentiate the following functions.
(a)
g
(
t
)
=
(
t
5
+
t
2
)
6
t
g(t)=(t^5+t^2)^{6t}
g
(
t
)
=
(
t
5
+
t
2
)
6
t
(b)
f
(
x
)
=
x
6
x
f(x)=\sqrt x^{6x}
f
(
x
)
=
x
6
x
(c)
f
(
x
)
=
(
x
3
+
4
)
tan
x
f(x)=(x^3+4)^{\tan x}
f
(
x
)
=
(
x
3
+
4
)
t
a
n
x
(d) Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
if
y
cos
(
x
)
=
x
tan
(
y
)
y\cos(x)=x\tan(y)
y
cos
(
x
)
=
x
tan
(
y
)
Differentiate
y
with respect to
x
in the equation
2
x
2
+
3
x
y
−
y
2
=
1
2x^2+3xy-y^2=1
2
x
2
+
3
x
y
−
y
2
=
1
.
Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
given that
e
x
y
−
x
y
=
2
x
2
+
1
e^{xy}-xy=2x^2+1
e
x
y
−
x
y
=
2
x
2
+
1
Evaluate the slope of the tangent line to the curve at the given point.
x
3
+
y
3
=
4
x
y
,
at
P
(
2
,
2
)
x^3+y^3=4xy ,\,\, \text{at} \,\, P(2,2)
x
3
+
y
3
=
4
x
y
,
at
P
(
2
,
2
)
Evaluate the slope of the tangent line to the curve at the given point.
x
3
+
y
3
=
y
sin
x
+
1
,
at
P
(
0
,
1
)
x^3+y^3=y\sin{x}+1,\,\, \text{at}\,\,P(0,1)
x
3
+
y
3
=
y
sin
x
+
1
,
at
P
(
0
,
1
)
Implicit Differentiation
Given
x
2
+
y
=
y
2
−
4
x^2+y=y^2-4
x
2
+
y
=
y
2
−
4
, find
y
′
y'
y
′
.
Implicit Differentiation
Find
d
y
d
x
\displaystyle \frac{\text{d}y}{\text{d}x}
d
x
d
y
for
y
3
=
2
x
y^3=2x
y
3
=
2
x
.
Find the point(s) where
x
2
−
4
x
+
6
y
+
y
2
=
−
9
x^2-4x+6y+y^2 = −9
x
2
−
4
x
+
6
y
+
y
2
=
−
9
has horizontal and vertical tangent lines.
Find
d
y
d
x
\dfrac{dy}{dx}
d
x
d
y
for the function
y
cos
x
=
x
2
+
cos
y
y\cos x=x^2+\cos y
y
cos
x
=
x
2
+
cos
y
Logarithmic for Implicitly Defined Function
Find
d
y
/
d
x
dy/dx
d
y
/
d
x
at the point
(
1
,
1
)
(1,1)
(
1
,
1
)
given
x
y
=
y
x
x^y=y^x
x
y
=
y
x
Implicit Second Derivative
Find
d
2
y
d
x
2
\dfrac{d^2y}{dx^2}
d
x
2
d
2
y
at the point
(
2
,
2
)
(2,2)
(
2
,
2
)
given
x
y
+
y
2
=
2
xy+y^2=2
x
y
+
y
2
=
2
Practice: Tangent and Normal Lines with Implicit
Q:
\textbf{Q:}
Q:
Find the equation of the tangent and the normal lines to the curve
x
2
cos
2
y
−
sin
y
=
0
x^2\cos^2y-\sin y=0
x
2
cos
2
y
−
sin
y
=
0
at the point
(
0
,
π
)
(0,\pi)
(
0
,
π
)
.
Implicit Differentiation
Evaluate the slope of the line tangent to the curve
x
y
=
x
3
y
−
6
\sqrt{xy}=x^3y-6
x
y
=
x
3
y
−
6
at the point
P
(
1
,
9
)
P(1,9)
P
(
1
,
9
)
.
Implicit with Trig
Find
d
y
/
d
x
dy/dx
d
y
/
d
x
for
x
y
=
tan
(
x
+
y
2
)
\displaystyle xy=\tan(x+y^{2})
x
y
=
tan
(
x
+
y
2
)
Find the slope of the tangent line to the curve
ln
(
x
−
y
)
+
3
y
2
=
3
\ln(x-y)+3y^2=3
ln
(
x
−
y
)
+
3
y
2
=
3
at point
(
3
,
2
)
(3,2)
(
3
,
2
)
.
Find
y
′
y'
y
′
for the function
x
4
+
x
3
y
−
x
y
2
=
1
x^4+x^3y-xy^2=1
x
4
+
x
3
y
−
x
y
2
=
1
Implicit Differentiation: Estimate of Implicit Function
Use a linear approximation to estimate the
y
y
y
value of the curve described by
x
3
+
3
x
y
+
y
3
=
5
x^3+3xy+y^3=5
x
3
+
3
x
y
+
y
3
=
5
at the point for which
x
=
0.1
x=0.1
x
=
0.1
(leave it as an exact value if you're not allowed a calculator on the exam).
Find the point on the graph
y
2
= 4
x
+ 5 where the tangent line is parallel to the line
y
= 2
x
+ 184.
Concept Clarifier
Compute the second derivative of the implicit function
y
3
=
x
cos
y
y^3=x \cos y
y
3
=
x
cos
y
, where
y
y
y
is a function of
x
x
x
.
Do not simplify.
Find the point(s) where
x
2
−4
x
+6
y
+
y
2
= −9 has horizontal and vertical tangent lines.
Find the slope of the tangent line to the curve ln(
x
−
y
) +
y
2
= 4 at point (3,2).
(Express your answer as a fraction in lowest terms. If the answer is a negative number, put the negative sign in front of the entire fraction.)
Find the equation of the tangent line at the point
(
1
,
2
)
(1,2)
(
1
,
2
)
of the curve
y
3
−
2
x
5
y
2
=
y
ln
(
x
)
y^3-2x^5y^2=y\ln(x)
y
3
−
2
x
5
y
2
=
y
ln
(
x
)
.
Find the derivative of
sin
(
x
+
y
)
−
s
e
c
(
x
2
+
y
2
)
=
y
\sin(x+y)-\mathrm{sec} (x^2+y^2)=y
sin
(
x
+
y
)
−
sec
(
x
2
+
y
2
)
=
y
.
Find the derivative of
y
with respect to
x
of the following equation,
y
2
+
x
5
y
3
=
5
−
ln
y
y^2+x^5y^3=5-\ln y
y
2
+
x
5
y
3
=
5
−
ln
y
.
Given that
x
2
+
y
=
y
2
−
4
x^2+y=y^2-4
x
2
+
y
=
y
2
−
4
, where
y
is dependent on
x
, find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
Implicit Differentiation
For the implicit equation
x
4
+
x
3
y
−
x
y
2
=
1
x^4+x^3y-xy^2=1
x
4
+
x
3
y
−
x
y
2
=
1
, where y =
f
(
x
), find
y
'.
Evaluate the slope of the tangent line to the curve at the given point.
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
\displaystyle \sqrt{xy}=x^3y-6\,\, at \,\,P(1,9)
x
y
=
x
3
y
−
6
a
t
P
(
1
,
9
)
Find the tangent to the curve 4𝑦
2
− 𝑒
𝑥
+ 𝑥 = 4𝑥 + 3 at the point where 𝑥 = 0 and 𝑦 is positive
Implicit Differentiation
Find all the equations of the tangent lines to the curve
y
=
x
3
+
x
2
−
x
y=x^3+x^2-x
y
=
x
3
+
x
2
−
x
that are parallel to the x-axis.
Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
at
x
=
0
x=0
x
=
0
if
x
cos
(
y
2
)
=
y
cos
x
x\cos\left(\frac{y}{2}\right)=y\cos x
x
cos
(
2
y
)
=
y
cos
x
.
Implicit Differentiation: Tangent line
Find the equation of the tangent to the graph
3
y
3
+
x
=
x
2
+
1
3y^3+x=x^2+1
3
y
3
+
x
=
x
2
+
1
at
x
=
2
x=2
x
=
2
.
Implicit differentiation
Differentiate
e
x
y
−
tan
y
=
3
x
3
+
2
x
e^{xy}-\tan y=\frac{3}{x^3}+2x
e
x
y
−
tan
y
=
x
3
3
+
2
x
Implicit differentiation
Suppose
3
x
3
+
4
y
2
=
19
3x^3+4y^2=19
3
x
3
+
4
y
2
=
19
, where
x
,
y
x,\ y
x
,
y
are both functions of
t
t
t
.
If
d
y
d
t
=
3
\frac{dy}{dt}=3
d
t
d
y
=
3
, find
d
x
d
t
\frac{dx}{dt}
d
t
d
x
when
y
=
2
y=2
y
=
2
.
Solve for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
:
x
2
y
+
3
y
2
x
2
=
x
+
y
x^2y+3y^2x^2=x+y
x
2
y
+
3
y
2
x
2
=
x
+
y
Implicit Differentiation
Use implicit differentiation to calculate the derivative of
θ
(
x
)
=
arctan
(
x
)
\theta(x) = \arctan(x)
θ
(
x
)
=
arctan
(
x
)
Implicit Differentiation
Find
d
y
d
x
\displaystyle \frac{\text{d}y}{\text{d}x}
d
x
d
y
for
ln
(
x
2
+
y
2
)
+
2
x
y
=
4
\ln(x^2+y^2)+2xy=4
ln
(
x
2
+
y
2
)
+
2
x
y
=
4
.
Evaluate the slope of the tangent line to the curve at the given point.
x
2
y
3
+
x
ln
y
+
y
sin
x
=
10
a
t
P
(
x
,
y
)
\displaystyle x^2y^3+x\ln{y}+y\sin{x}=10\,\, at \,\,P(x,y)
x
2
y
3
+
x
ln
y
+
y
sin
x
=
10
a
t
P
(
x
,
y
)
Practice: Implicit Differentiation
Find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
for the following
tan
y
x
=
x
\tan{\frac{y}{x}}=x
tan
x
y
=
x