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Find the point on the graph y2 = 4x + 5 where the tangent line is parallel to t…
Related Topics
Wize University Calculus 1 Textbook > Derivatives
Implicit Differentiation
4 Activities
Wize University Calculus 1 Textbook > Derivatives
Tangent Lines
3 Activities
Find the point on the graph
y
2
= 4
x
+ 5 where the tangent line is parallel to the line
y
= 2
x
+ 184.
(
1
,
1
)
(1,1)
(
1
,
1
)
(
1
,
−
1
)
(1,-1)
(
1
,
−
1
)
(
−
1
,
1
)
(-1,1)
(
−
1
,
1
)
(
−
1
,
−
1
)
(-1,-1)
(
−
1
,
−
1
)
I don't know
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).
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
.
Implicit Differentiation
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
)
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
More Tangent Lines Questions:
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
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
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
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
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for
0
≤
t
≤
4
0\le t\le4
0
≤
t
≤
4
. s(t) represents the position of the particle at time t.
Tangent Lines
Find the point
(
a
,
b
)
\left(a,b\right)
(
a
,
b
)
at which the tangent line to the curve
y
=
2
x
2
+
3
x
−
7
y=2x^2+3x-7
y
=
2
x
2
+
3
x
−
7
is parallel to the tangent line of the curve
y
=
3
x
4
−
5
x
+
1
y=3x^4-5x+1
y
=
3
x
4
−
5
x
+
1
at the point
(
1
,
−
6
)
\left(1,-6\right)
(
1
,
−
6
)
.
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \dfrac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \dfrac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
Tangent Lines
Is there any point on the graph of
y
=
x
2
+
3
x
+
8
y = x^2+ 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with the equation
y
=
7
x
+
104
y = 7x + 104
y
=
7
x
+
104
?
Tangent Lines
The tangent line to the curve
y
=
−
x
2
−
3
y=-x^2-3
y
=
−
x
2
−
3
at
x
=
a
x=a
x
=
a
passes through
(
0
,
1
)
(0, 1)
(
0
,
1
)
. Find
a
a
a
and the corresponding function value.
Is there any point on the graph
y
=
x
2
+
3
x
+
8
y=x^2 + 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with equation
y
=
5
x
+
102
y = 5x + 102
y
=
5
x
+
102
?
Practice: Tangent Through Outside Point
Q:
\textbf{Q:}
Q:
There are points on the curve
y
=
2
−
x
4
+
x
y=\dfrac{2-x}{4+x}
y
=
4
+
x
2
−
x
at which the tangent line passes through the origin. Find all such points.
Tangent Line Intercept
Let
f
(
x
)
=
3
x
2
−
4
x
+
9
f(x)=3x^2-4x+9
f
(
x
)
=
3
x
2
−
4
x
+
9
and let L denote the tangent line to the graph of
f
(
x
)
f(x)
f
(
x
)
at the point on the graph where
x
=
2
x=2
x
=
2
. Find the point where L intersects the x-axis.
Tangent Lines
Consider the line
y
=
4
x
+
3
y = 4x + 3
y
=
4
x
+
3
. To which of the following functions is it tangent at
x
=
1.
x = 1.
x
=
1.
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
Consider the function
f
(
x
)
=
1
x
2
f(x) = \frac{1}{x^2}
f
(
x
)
=
x
2
1
. Let
g
(
a
)
g(a)
g
(
a
)
be the
x
x
x
- intercept of the tangent line of
f
(
x
)
f(x)
f
(
x
)
at
x
=
a
x = a
x
=
a
. Find
g
(
a
)
g(a)
g
(
a
)
for all
a
≠
0
a \not = 0
a
=
0
.
Given the function
f
(
x
)
=
x
3
−
3
x
+
1
f(x) = x^3 - 3x + 1
f
(
x
)
=
x
3
−
3
x
+
1
, find the equation of the tangent line at
x
=
2
x = 2
x
=
2
and find all other points that have tangent lines that are parallel to the one at
x
=
2
x = 2
x
=
2
.
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
)
Derivatives and Tangent Lines
A certain particle is traveling through space and time. We are able to measure the position of the particle at certain time values. Although we can't measure the particle continuously, we know that the particles position function is differentiable. The table gives values of time for the differentiable function s(t) for
0
≤
t
≤
4
0\le t\le4
0
≤
t
≤
4
. s(t) represents the position of the particle at time t.
Finding Tangent and Normal Lines
Find the tangent line and normal line to
f
(
x
)
=
2
x
2
+
x
−
1
f(x)=2x^2+x-1
f
(
x
)
=
2
x
2
+
x
−
1
at
x
=
1
x=1
x
=
1
.
Tangent Lines
Let
f
(
x
)
=
1
3
x
3
+
1
2
x
2
+
5
x
−
4
f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2+5x-4
f
(
x
)
=
3
1
x
3
+
2
1
x
2
+
5
x
−
4
, and suppose
g(x)
is its inverse function. Find the
x
-coordinates of all points on the graph of
g(x)
so that the tangent line at those points are perpendicular to the function
y
=
−
7
x
+
2.
y=-7x+2.
y
=
−
7
x
+
2.
Tangent Lines
Find a horizontal tangent line of the curve
f
(
x
)
=
x
3
/
3
−
2
x
2
−
5
x
+
4
f(x)=x^3/3-2x^2-5x+4
f
(
x
)
=
x
3
/3
−
2
x
2
−
5
x
+
4
Tangent Lines
Find all points on the graph of
f
(
x
)
=
2
x
2
+
x
−
3
f(x)=2x^2+x-3
f
(
x
)
=
2
x
2
+
x
−
3
whose tangent line passes through the point (3,0).
In the function
C
=
−
4
k
t
+
3
C=-4kt+3
C
=
−
4
k
t
+
3
, where
C
C
C
represents the concentration of a certain substance in a tank of water,
t
t
t
represents time, and
k
k
k
is a positive constant, the rate of change of the dependent variable with respect to the independent variable is a constant. The statement is:
Find the equation of tangent line to
y
=
2
x
+
3
y=\sqrt{2x+3}
y
=
2
x
+
3
at
x
=
3
x=3
x
=
3
. Express in slope-intercept form.
Is there any point on the graph
y
=
x
2
+
3
x
+
8
y=x^2 + 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with equation
y
=
5
x
+
102
y = 5x + 102
y
=
5
x
+
102
?
Tangent Lines
Find the horizontal tangent line of the curve
f
(
x
)
=
x
2
3
−
2
x
2
−
5
x
+
4
\displaystyle f(x)=\frac{x^2}{3}-2x^2-5x+4
f
(
x
)
=
3
x
2
−
2
x
2
−
5
x
+
4
Tangent Lines
Is there any point on the graph of
y
=
x
2
+
3
x
+
8
y = x^2+ 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with the equation
y
=
7
x
+
104
y = 7x + 104
y
=
7
x
+
104
?
Tangent Lines
The tangent line to the curve
y
=
−
x
2
−
3
y=-x^2-3
y
=
−
x
2
−
3
at
x
=
a
x=a
x
=
a
passes through
(
0
,
1
)
(0, 1)
(
0
,
1
)
. Find
a
a
a
and the corresponding function value.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \dfrac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
Practice: Tangent Through Outside Point
Q:
\textbf{Q:}
Q:
There are points on the curve
y
=
2
−
x
4
+
x
y=\dfrac{2-x}{4+x}
y
=
4
+
x
2
−
x
at which the tangent line passes through the origin. Find all such points.
Tangent Line Intercept
Let
f
(
x
)
=
3
x
2
−
4
x
+
9
f(x)=3x^2-4x+9
f
(
x
)
=
3
x
2
−
4
x
+
9
and let L denote the tangent line to the graph of
f
(
x
)
f(x)
f
(
x
)
at the point on the graph where
x
=
2
x=2
x
=
2
. Find the point where L intersects the x-axis.
Tangent Lines
At which point on the graph
y
=
2
x
+
5
y=\sqrt{2x+5}
y
=
2
x
+
5
the tangent line is parallel to the line with the equation
y
=
1
3
x
+
11
?
y=\frac{1}{3}x+11 ?
y
=
3
1
x
+
11
?
If
f
(
4
)
=
1
f(4)=1
f
(
4
)
=
1
,
f
′
(
4
)
=
−
1
f'(4)=-1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x)=\frac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
, then find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
How many distinct horizontal lines are tangent to the equation
f
(
x
)
=
x
2
(
9
−
x
2
)
f\left(x\right)=x^2\left(9-x^2\right)
f
(
x
)
=
x
2
(
9
−
x
2
)
?
🌶️
TOUGH!
As shown in the figure below, the tangent line to the graph
y
=
f
(
x
) intersects the
x
-axis at
x
=
b
. What is the value of
b
in terms of
a
,
f
(
a
), and
f
’(
a
)?
Find the point where the tangent line is horizontal for the curve
f
(
x
)
=
x
2
3
−
2
x
2
−
5
x
+
4
f(x)=\frac{x^2}{3}-2x^2-5x+4
f
(
x
)
=
3
x
2
−
2
x
2
−
5
x
+
4
.
Find the
y
y
y
-intercept of the tangent line of
f
(
x
)
=
1
x
+
x
at
x
=
2
f(x)=\frac{1}{\sqrt{x}}+\sqrt{x} \ \ \text{at }\ x=2
f
(
x
)
=
x
1
+
x
at
x
=
2
.
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \frac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
then find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
.
🌶️
TOUGH!
Let
f
(
x
)
=
x
2
f(x)=x^2
f
(
x
)
=
x
2
, given the tangent line of
f
f
f
at
x
=
a
x=a
x
=
a
passes through the point
(
4
,
15
)
(4,15)
(
4
,
15
)
not on the function, find
a
a
a
.
Find the equation of the tangent line to
f
(
x
)
=
(
x
2
−
2
)
sin
x
+
2
x
cos
x
f(x) = (x^2 - 2) \sin x+2x \cos x
f
(
x
)
=
(
x
2
−
2
)
sin
x
+
2
x
cos
x
at
x
=
π
x = \pi
x
=
π
.
Practice: Equation of tangent (and normal) line
Find the tangent and normal lines to the graph
f
(
x
)
=
x
f\left(x\right)=\sqrt{x}
f
(
x
)
=
x
at
x = 9
. Which point on this graph is the slope of the tangent line parallel to the line
y
=
3
x
+
1
?
y = 3x + 1?
y
=
3
x
+
1
?
Find the equation of the tangent line to the given point for the following function.
f
(
x
)
=
e
x
2
+
x
,
at
P
(
1
,
e
2
)
f(x)=e^{x^2+\sqrt{x}},\,\text{at}\,P(1,e^2)
f
(
x
)
=
e
x
2
+
x
,
at
P
(
1
,
e
2
)
Tangent Lines
Find the point
(
a
,
b
)
\left(a,b\right)
(
a
,
b
)
at which the tangent line to the curve
y
=
2
x
2
+
3
x
−
7
y=2x^2+3x-7
y
=
2
x
2
+
3
x
−
7
is parallel to the tangent line of the curve
y
=
3
x
4
−
5
x
+
1
y=3x^4-5x+1
y
=
3
x
4
−
5
x
+
1
at the point
(
1
,
−
6
)
\left(1,-6\right)
(
1
,
−
6
)
.
Tangent Lines
If
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
and
f
′
(
4
)
=
−
1
f'(4) = -1
f
′
(
4
)
=
−
1
and
g
(
x
)
=
f
(
x
)
x
−
3
g(x) = \frac{f(x)}{x-3}
g
(
x
)
=
x
−
3
f
(
x
)
then find the equation of the tangent line to the curve
g
(
x
)
g(x)
g
(
x
)
at
x
=
4
x=4
x
=
4
?
Is there any point on the graph
y
=
x
2
+
3
x
+
8
y=x^2 + 3x + 8
y
=
x
2
+
3
x
+
8
such that the tangent line is parallel to the line with the equation
y
=
5
x
+
102
y = 5x + 102
y
=
5
x
+
102
?
Slope of a Tangent Line
Practice: Slope of a Tangent Line
Find all values of
x
x
x
such that the tangent line to the curve
y
=
2
x
3
−
x
2
2
−
x
−
5
y=2x^3-\frac{x^2}{2}-x-5
y
=
2
x
3
−
2
x
2
−
x
−
5
is horizontal.
Solving for the tangent line
Let
f
(
x
)
=
x
2
+
a
x
−
3
f(x) = x^2 + ax - 3
f
(
x
)
=
x
2
+
a
x
−
3
be a function such that at
x
=
2
x = 2
x
=
2
, the tangent line is parallel to the line given by
y
=
x
+
3
y = x + 3
y
=
x
+
3
. Find what
a
a
a
is and give the equation of the tangent line.
Tangent Lines
Consider the function
f
(
x
)
=
−
2
x
2
+
3
x
−
1
f(x) = - 2x^2 + 3x - 1
f
(
x
)
=
−
2
x
2
+
3
x
−
1
. There is a point
(
a
,
b
)
(a, b)
(
a
,
b
)
such that the tangent line to
f
f
f
at
(
a
,
b
)
(a, b)
(
a
,
b
)
is perpendicular to the line
y
=
x
3
+
5
y = \frac{x}{3} + 5
y
=
3
x
+
5
. Find the point
(
a
,
b
)
(a, b)
(
a
,
b
)
and write down the equation of the tangent line.
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
)