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The Quotient and Chain Rules
Related Topics
Wize University Calculus 1 Textbook > Derivatives
The Chain Rule
3 Activities
Wize University Calculus 1 Textbook > Derivatives
The Quotient Rule
3 Activities
Let
f
(
x
)
=
x
2
−
6
x
−
3
\displaystyle f(x) = \frac{\sqrt{x^2 - 6}}{x - 3}
f
(
x
)
=
x
−
3
x
2
−
6
.
Part 1
Part 2
Part 3
Compute and simplify
f
′
(
x
)
f'(x)
f
′
(
x
)
.
A)
−
3
x
+
6
(
x
−
3
)
2
x
2
−
6
\displaystyle \frac{-3x +6}{(x - 3)^2\sqrt{x^2 - 6}}
(
x
−
3
)
2
x
2
−
6
−
3
x
+
6
B)
x
x
2
−
6
\displaystyle \frac{x}{\sqrt{x^2 - 6}}
x
2
−
6
x
C)
−
3
x
+
6
(
x
−
3
)
2
\displaystyle \frac{-3x +6}{(x - 3)^2}
(
x
−
3
)
2
−
3
x
+
6
D)
x
(
x
−
3
)
2
x
2
−
6
\displaystyle \frac{x(x - 3)^2}{\sqrt{x^2 - 6}}
x
2
−
6
x
(
x
−
3
)
2
I don't know
Previous Part
Check Part 1
Next Part
Check Submission
More The Chain Rule Questions:
The Chain Rule
Q
:
\bf{Q:}
Q
:
Find
(
3
3
x
−
x
4
+
2
cos
−
1
(
x
)
−
2025
)
′
(3^{3x-x^{4}}+2^{\cos^{-1}(x)}-2025)'
(
3
3
x
−
x
4
+
2
c
o
s
−
1
(
x
)
−
2025
)
′
Derivatives
Find the derivative of the following functions at the given values
a)
f
(
x
)
=
arcsin
(
x
2
)
x
3
−
1
f\left(x\right)=\frac{\arcsin\left(x^2\right)}{x^3-1}
f
(
x
)
=
x
3
−
1
a
r
c
s
i
n
(
x
2
)
at
x
=
0
x=0
x
=
0
b)
g
(
x
)
=
(
cos
x
)
e
x
g\left(x\right)=\left(\cos x\right)^{e^x}
g
(
x
)
=
(
cos
x
)
e
x
at
x
=
0
x=0
x
=
0
Finding a Derivative
If
f
(
x
)
=
tan
−
1
(
3
sin
x
)
f(x)=\tan^{-1}\left(3^{\sin x}\right)
f
(
x
)
=
tan
−
1
(
3
s
i
n
x
)
, find
f
′
(
π
)
f'(\pi)
f
′
(
π
)
.
The Chain Rule
Q
:
\bf{Q:}
Q
:
Find
(
3
3
x
−
x
4
+
2
c
o
s
−
1
(
x
)
−
2019
)
′
(3^{3x-x^4}+2^{cos^{-1}(x)}-2019)'
(
3
3
x
−
x
4
+
2
co
s
−
1
(
x
)
−
2019
)
′
The Chain Rule
Find the derivative of
h
(
x
)
=
f
(
g
(
x
)
)
h(x)=f(g(x))
h
(
x
)
=
f
(
g
(
x
))
if
f
(
x
)
=
x
2
+
1
f(x)=x^2+1
f
(
x
)
=
x
2
+
1
and
g
(
x
)
=
3
x
g(x)=3\sqrt{x}
g
(
x
)
=
3
x
.
The Chain Rule
Find the derivative of
h
(
x
)
=
f
(
g
(
x
)
)
h(x)=f(g(x))
h
(
x
)
=
f
(
g
(
x
))
if
f
(
x
)
=
x
2
+
1
f(x)=x^2+1
f
(
x
)
=
x
2
+
1
and
g
(
x
)
=
3
x
g(x)=3\sqrt{x}
g
(
x
)
=
3
x
.
The Chain Rule
Find
g
′
(
0
)
g'(0)
g
′
(
0
)
for
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
,
g(x)=\sqrt{f(x^2)} + f(x),
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
,
given
f
(
0
)
=
2
and
f
′
(
0
)
=
1.
f(0)=2 \text{ and }f'(0)=1.
f
(
0
)
=
2
and
f
′
(
0
)
=
1.
The Chain Rule
If
2
−
g
(
x
)
=
x
2
+
2
[
f
(
x
)
]
2
−
x
3
g
(
x
)
2-g\left(x\right)=x^2+2\left[f\left(x\right)\right]^2-x^3g\left(x\right)
2
−
g
(
x
)
=
x
2
+
2
[
f
(
x
)
]
2
−
x
3
g
(
x
)
,
f
(
0
)
=
1
,
f
′
(
0
)
=
−
2
,
g
(
0
)
=
−
1
and
g
′
′
(
0
)
=
3
f\left(0\right)=1\ ,\ f'\left(0\right)=-2,\ g\left(0\right)=-1\ \text{and}\ \ g''\left(0\right)=3
f
(
0
)
=
1
,
f
′
(
0
)
=
−
2
,
g
(
0
)
=
−
1
and
g
′′
(
0
)
=
3
, then the value of
f
′
′
(
0
)
f''\left(0\right)
f
′′
(
0
)
is equal to
Find
g
′
(
0
)
g^{\prime}(0)
g
′
(
0
)
for
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
g(x)=\sqrt{f(x^2)}+f(x)
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
, given
f
(
0
)
=
2
f(0)=2
f
(
0
)
=
2
and
f
′
(
0
)
=
1
f^{\prime}(0)=1
f
′
(
0
)
=
1
.
The Chain Rule
Find the derivative of the following function.
f
(
x
)
=
(
2
x
+
1
x
2
+
1
)
3
\displaystyle f(x)=\left(\frac{2x+1}{x^2+1}\right)^3
f
(
x
)
=
(
x
2
+
1
2
x
+
1
)
3
Find
d
d
x
(
arctan
(
e
tan
x
)
)
\frac{d}{dx}\left(\arctan\left(e^{\tan x}\right)\right)
d
x
d
(
arctan
(
e
t
a
n
x
)
)
.
The Chain Rule
The derivative of
y
=
3
e
x
y=3^{e^x}
y
=
3
e
x
is
The Chain Rule
The derivative of
y
=
3
e
x
y=3^{e^x}
y
=
3
e
x
is
The Chain Rule
The derivative of
y
=
3
e
x
y=3^{e^x}
y
=
3
e
x
is
The Chain Rule
The derivative of
y
=
3
e
x
y=3^{e^x}
y
=
3
e
x
is
Practice: Chain Rule
Find the derivative of
y
=
ln
(
arctan
x
)
y=\ln(\arctan x)
y
=
ln
(
arctan
x
)
Find
d
d
x
(
arctan
(
e
tan
x
)
)
\frac{d}{dx}\left(\arctan\left(e^{\tan x}\right)\right)
d
x
d
(
arctan
(
e
t
a
n
x
)
)
.
The Chain Rule: Finding a derivative
Find the derivative of
f
(
x
)
=
cos
(
e
2
x
)
f\left(x\right)=\sqrt{\cos\left(e^{2x}\right)}
f
(
x
)
=
cos
(
e
2
x
)
Practice: trig and inverse trig derivatives
Practice: trig and inverse trig derivatives
Find the derivative of the following trigonometric and inverse trigonometric functions:
a)
f
(
x
)
=
arcsin
(
3
x
2
)
f(x)=\arcsin(3x^2)
f
(
x
)
=
arcsin
(
3
x
2
)
The Chain Rule
Given this table of values for
f
,
g
,
f
′
,
and
g
′
f,\ g,\ f',\ \text{and}\ g'
f
,
g
,
f
′
,
and
g
′
below, answer the following questions.
x
f
(
x
)
f
′
(
x
)
f
′
′
(
x
)
g
(
x
)
g
′
(
x
)
0
0
−
1
−
5
2
3
π
2
π
1
0
4
5
2
−
2
−
4
10
π
2
−
3
\begin{array}{|c|c|c|c|c|c|} \hline x&f(x)&f'(x)&f''(x)&g(x)&g'(x)\\ \hline 0&0&-1&-5&2&3\\ \hline \frac{\pi}{2}&\pi&1&0&4&5\\ \hline 2&-2&-4&10&\frac{\pi}{2}&-3\\ \hline \end{array}
x
0
2
π
2
f
(
x
)
0
π
−
2
f
′
(
x
)
−
1
1
−
4
f
′′
(
x
)
−
5
0
10
g
(
x
)
2
4
2
π
g
′
(
x
)
3
5
−
3
The Chain Rule
Q
:
\bf{Q:}
Q
:
Find
(
3
3
x
−
x
4
+
2
c
o
s
−
1
(
x
)
−
2019
)
′
(3^{3x-x^4}+2^{cos^{-1}(x)}-2019)'
(
3
3
x
−
x
4
+
2
co
s
−
1
(
x
)
−
2019
)
′
The Chain Rule
Find the derivative of
h
(
x
)
=
f
(
g
(
x
)
)
h(x)=f(g(x))
h
(
x
)
=
f
(
g
(
x
))
if
f
(
x
)
=
x
2
+
1
f(x)=x^2+1
f
(
x
)
=
x
2
+
1
and
g
(
x
)
=
3
x
g(x)=3\sqrt{x}
g
(
x
)
=
3
x
.
Practice: Chain Rule with Given Values
Q.
\textbf{Q.}
Q.
Given that
g
(
1
)
=
g
′
(
1
)
=
1
g(1)=g^{\prime}(1)=1
g
(
1
)
=
g
′
(
1
)
=
1
, find
f
′
(
π
/
2
)
f^{\prime}(\pi/2)
f
′
(
π
/2
)
if
f
(
x
)
=
g
(
sin
x
)
x
+
1
\displaystyle f(x)=\frac{\sqrt{g(\sin{x})}}{x+1}
f
(
x
)
=
x
+
1
g
(
sin
x
)
.
Practice: Chain Rule with Given Values
Q:
\textbf{Q:}
Q:
Given that
f
(
1
)
=
5
,
g
(
1
)
=
−
2
,
f
′
(
1
)
=
2
f(1)=5,\ g(1)=-2,\ f'(1)=2
f
(
1
)
=
5
,
g
(
1
)
=
−
2
,
f
′
(
1
)
=
2
and
g
′
(
1
)
=
−
1
/
2
g'(1)=-1/2
g
′
(
1
)
=
−
1/2
, find the derivative of
f
(
x
)
−
g
2
(
x
)
\sqrt{f(x)-g^2(x)}
f
(
x
)
−
g
2
(
x
)
at
x
=
1
x=1
x
=
1
.
The chain rule
Find the derivative of
h
(
x
)
=
log
(
cos
(
x
)
)
h(x) = \log(\cos(x))
h
(
x
)
=
lo
g
(
cos
(
x
))
. 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
.
Derivatives
Find the derivative of the following functions at the given values
a)
f
(
x
)
=
arcsin
(
x
2
)
x
3
−
1
f\left(x\right)=\frac{\arcsin\left(x^2\right)}{x^3-1}
f
(
x
)
=
x
3
−
1
a
r
c
s
i
n
(
x
2
)
at
x
=
0
x=0
x
=
0
b)
g
(
x
)
=
(
cos
x
)
e
x
g\left(x\right)=\left(\cos x\right)^{e^x}
g
(
x
)
=
(
cos
x
)
e
x
at
x
=
0
x=0
x
=
0
The chain rule: Logarithmic Derivatives
The derivative of
log
3
(
e
3
x
)
\log_3\left(e^{3x}\right)
lo
g
3
(
e
3
x
)
is
The chain rule
The derivative of
y
=
(
2
−
x
7
)
500
y=\left(2-x^7\right)^{500}
y
=
(
2
−
x
7
)
500
is
The Chain Rule
If
2
−
g
(
x
)
=
x
2
+
2
[
f
(
x
)
]
2
−
x
3
g
(
x
)
2-g\left(x\right)=x^2+2\left[f\left(x\right)\right]^2-x^3g\left(x\right)
2
−
g
(
x
)
=
x
2
+
2
[
f
(
x
)
]
2
−
x
3
g
(
x
)
,
f
(
0
)
=
1
,
f
′
(
0
)
=
−
2
,
g
(
0
)
=
−
1
and
g
′
′
(
0
)
=
3
f\left(0\right)=1\ ,\ f'\left(0\right)=-2,\ g\left(0\right)=-1\ \text{and}\ \ g''\left(0\right)=3
f
(
0
)
=
1
,
f
′
(
0
)
=
−
2
,
g
(
0
)
=
−
1
and
g
′′
(
0
)
=
3
, then the value of
f
′
′
(
0
)
f''\left(0\right)
f
′′
(
0
)
is equal to
Find the derivative of
h
(
x
)
=
f
(
g
(
x
)
)
if
f
(
x
)
=
x
2
+
1
h(x)=f(g(x))\,\, \text{ if }\, f(x)=x^2+1
h
(
x
)
=
f
(
g
(
x
))
if
f
(
x
)
=
x
2
+
1
and
g
(
x
)
=
3
x
g(x)=3\sqrt{x}
g
(
x
)
=
3
x
.
Find
g
′
(
0
)
g^{\prime}(0)
g
′
(
0
)
for
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
g(x)=\sqrt{f(x^2)}+f(x)
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
, given
f
(
0
)
=
2
f(0)=2
f
(
0
)
=
2
and
f
′
(
0
)
=
1
f^{\prime}(0)=1
f
′
(
0
)
=
1
.
Find the derivative of
h
(
x
)
=
f
(
g
(
x
)
)
if
f
(
x
)
=
x
2
+
1
h(x)=f(g(x))\,\, \text{ if }\, f(x)=x^2+1
h
(
x
)
=
f
(
g
(
x
))
if
f
(
x
)
=
x
2
+
1
and
g
(
x
)
=
3
x
g(x)=3\sqrt{x}
g
(
x
)
=
3
x
.
Find the derivative of the following function.
f
(
x
)
=
(
2
x
+
1
x
2
+
1
)
3
\displaystyle f(x)=\left(\frac{2x+1}{x^2+1}\right)^3
f
(
x
)
=
(
x
2
+
1
2
x
+
1
)
3
The Chain Rule
Find
g
′
(
0
)
g'(0)
g
′
(
0
)
for
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
,
g(x)=\sqrt{f(x^2)} + f(x),
g
(
x
)
=
f
(
x
2
)
+
f
(
x
)
,
given
f
(
0
)
=
2
and
f
′
(
0
)
=
1.
f(0)=2 \text{ and }f'(0)=1.
f
(
0
)
=
2
and
f
′
(
0
)
=
1.
The Chain Rule
Compute the derivative of
f
(
x
)
=
x
4
+
4
x
4
+
4
f(x) = \sqrt{x^4 + \frac{4}{x^4}+4}
f
(
x
)
=
x
4
+
x
4
4
+
4
Practice: Chain Rule
Find the derivative of the following function:
y
=
sin
(
x
)
y=\sin(\sqrt{x})
y
=
sin
(
x
)
Practice: Complicated Chain-Rule
Find an expression for the derivative of the following function:
f
(
x
)
=
cos
(
e
2
x
)
f\left(x\right)=\sqrt{\cos \left(e^{2x}\right)}
f
(
x
)
=
cos
(
e
2
x
)
Practice: Chain Rule
Find the derivative of the following function:
y
=
e
sin
(
x
)
y=e^{\sin(\sqrt{x})}
y
=
e
s
i
n
(
x
)
Find the derivative of the following function.
f
(
x
)
=
arcsinx
\displaystyle f(x)=\sqrt{\text{arcsin{x}}}
f
(
x
)
=
arcsin
x
The Chain Rule
Q
:
\bf{Q:}
Q
:
Find
(
3
3
x
−
x
4
+
2
c
o
s
−
1
(
x
)
−
2019
)
′
(3^{3x-x^4}+2^{cos^{-1}(x)}-2019)'
(
3
3
x
−
x
4
+
2
co
s
−
1
(
x
)
−
2019
)
′
The Chain Rule
Find the derivative of
h
(
x
)
=
f
(
g
(
x
)
)
h(x)=f(g(x))
h
(
x
)
=
f
(
g
(
x
))
if
f
(
x
)
=
x
2
+
1
f(x)=x^2+1
f
(
x
)
=
x
2
+
1
and
g
(
x
)
=
3
x
g(x)=3\sqrt{x}
g
(
x
)
=
3
x
.
Practice: Chain Rule with Given Values
Q.
\textbf{Q.}
Q.
Given that
g
(
1
)
=
g
′
(
1
)
=
1
g(1)=g^{\prime}(1)=1
g
(
1
)
=
g
′
(
1
)
=
1
, find
f
′
(
π
/
2
)
f^{\prime}(\pi/2)
f
′
(
π
/2
)
if
f
(
x
)
=
g
(
sin
x
)
x
+
1
\displaystyle f(x)=\frac{\sqrt{g(\sin{x})}}{x+1}
f
(
x
)
=
x
+
1
g
(
sin
x
)
.
Practice: Chain Rule with Given Values
Q:
\textbf{Q:}
Q:
Given that
f
(
1
)
=
5
,
g
(
1
)
=
−
2
,
f
′
(
1
)
=
2
f(1)=5,\ g(1)=-2,\ f'(1)=2
f
(
1
)
=
5
,
g
(
1
)
=
−
2
,
f
′
(
1
)
=
2
and
g
′
(
1
)
=
−
1
/
2
g'(1)=-1/2
g
′
(
1
)
=
−
1/2
, find the derivative of
f
(
x
)
−
g
2
(
x
)
\sqrt{f(x)-g^2(x)}
f
(
x
)
−
g
2
(
x
)
at
x
=
1
x=1
x
=
1
.
Given the information
g
(
x
)
=
1
+
[
f
(
x
)
]
2
such that
f
(
1
)
=
2
,
f
′
(
1
)
=
−
3
g(x)=\sqrt{1+[f(x)]^2} \ \text{ such that } f(1)=2,f'(1)=-3
g
(
x
)
=
1
+
[
f
(
x
)
]
2
such that
f
(
1
)
=
2
,
f
′
(
1
)
=
−
3
. Find
g
′
(
1
)
.
g'(1).
g
′
(
1
)
.
Express your answer as a fraction in lowest terms. If the answer is negative, put the negative sign in front of the entire fraction.
The Chain Rule" Derivatives of Trigonometric Functions
Calculate the derivative of the function
f
(
x
)
=
tan
(
x
3
)
f\left(x\right)=\sqrt{\tan\left(x^3\right)}
f
(
x
)
=
tan
(
x
3
)
The Chain Rule
If
g
(
t
)
=
9
+
f
(
t
)
3
g\left(t\right)=\sqrt[3]{9+f(t)}
g
(
t
)
=
3
9
+
f
(
t
)
, write an expression for the derivative
g
′
(
t
)
g'\left(t\right)
g
′
(
t
)
The Chain Rule
Find the derivative of the following function.
f
(
x
)
=
(
2
x
+
1
x
2
+
1
)
3
\displaystyle f(x)=\left(\frac{2x+1}{x^2+1}\right)^3
f
(
x
)
=
(
x
2
+
1
2
x
+
1
)
3
Practice: Chain Rule
Find the derivative of
y
=
ln
(
arctan
x
)
y=\ln(\arctan x)
y
=
ln
(
arctan
x
)
Find
d
d
x
(
arctan
(
e
tan
x
)
)
\frac{d}{dx}\left(\arctan\left(e^{\tan x}\right)\right)
d
x
d
(
arctan
(
e
t
a
n
x
)
)
.
Practice: trig and inverse trig derivatives
Practice: trig and inverse trig derivatives
Find the derivative of the following trigonometric and inverse trigonometric functions:
a)
f
(
x
)
=
arcsin
(
3
x
2
)
f(x)=\arcsin(3x^2)
f
(
x
)
=
arcsin
(
3
x
2
)
The Chain Rule: Finding a derivative
Find the derivative of
f
(
x
)
=
cos
(
e
2
x
)
f\left(x\right)=\sqrt{\cos\left(e^{2x}\right)}
f
(
x
)
=
cos
(
e
2
x
)
(Duplicated)
Find
f
′
(
1
)
f'\left(1\right)
f
′
(
1
)
if
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=\sqrt{x^2\ g\left(x\right)}
f
(
x
)
=
x
2
g
(
x
)
,
g
(
1
)
=
1
g\left(1\right)=1
g
(
1
)
=
1
and
g
′
(
1
)
=
2
g'\left(1\right)=2
g
′
(
1
)
=
2
.
(Duplicated)
If
f
(
x
)
=
(
g
(
x
)
−
h
(
x
)
)
5
f\left(x\right)=\left(g\left(x\right)-h\left(x\right)\right)^5
f
(
x
)
=
(
g
(
x
)
−
h
(
x
)
)
5
,
g
(
0
)
=
2
,
h
(
0
)
=
0
,
g
′
(
0
)
=
6
g\left(0\right)=2,\ h\left(0\right)=0,\ g'\left(0\right)=6
g
(
0
)
=
2
,
h
(
0
)
=
0
,
g
′
(
0
)
=
6
, and
h
′
(
0
)
=
−
4
h'\left(0\right)=-4
h
′
(
0
)
=
−
4
, find
f
′
(
0
)
f'\left(0\right)
f
′
(
0
)
.
The Chain Rule
The derivative of
y
=
3
e
x
y=3^{e^x}
y
=
3
e
x
is
The Chain Rule
Given this table of values for
f
,
g
,
f
′
,
and
g
′
f,\ g,\ f',\ \text{and}\ g'
f
,
g
,
f
′
,
and
g
′
below, answer the following questions.
x
f
(
x
)
f
′
(
x
)
f
′
′
(
x
)
g
(
x
)
g
′
(
x
)
0
0
−
1
−
5
2
3
π
2
π
1
0
4
5
2
−
2
−
4
10
π
2
−
3
\begin{array}{|c|c|c|c|c|c|} \hline x&f(x)&f'(x)&f''(x)&g(x)&g'(x)\\ \hline 0&0&-1&-5&2&3\\ \hline \frac{\pi}{2}&\pi&1&0&4&5\\ \hline 2&-2&-4&10&\frac{\pi}{2}&-3\\ \hline \end{array}
x
0
2
π
2
f
(
x
)
0
π
−
2
f
′
(
x
)
−
1
1
−
4
f
′′
(
x
)
−
5
0
10
g
(
x
)
2
4
2
π
g
′
(
x
)
3
5
−
3
Finding a Derivative
Practice: Finding a Derivative
(
tan
−
1
(
3
sin
x
)
)
′
∣
x
=
π
=
\left(\tan^{-1}\left(3^{\sin x}\right)\right)'|_{x=\pi}=
(
tan
−
1
(
3
s
i
n
x
)
)
′
∣
x
=
π
=
Practice: The Chain Rule
Find the derivative of the following function
y
=
e
sin
x
y=e^{\sin\sqrt{x}}
y
=
e
s
i
n
x
Derivatives: Trigonometric and Exponential Functions
Calculate the derivative of the following functions.
f
(
x
)
=
tan
(
arccos
(
e
4
x
)
)
\displaystyle f(x)=\tan(\text{arccos}(e^{4x}))
f
(
x
)
=
tan
(
arccos
(
e
4
x
))
More The Quotient Rule Questions:
Derivatives
Find the derivative of the following functions at the given values
a)
f
(
x
)
=
arcsin
(
x
2
)
x
3
−
1
f\left(x\right)=\frac{\arcsin\left(x^2\right)}{x^3-1}
f
(
x
)
=
x
3
−
1
a
r
c
s
i
n
(
x
2
)
at
x
=
0
x=0
x
=
0
b)
g
(
x
)
=
(
cos
x
)
e
x
g\left(x\right)=\left(\cos x\right)^{e^x}
g
(
x
)
=
(
cos
x
)
e
x
at
x
=
0
x=0
x
=
0
The quotient rule
Is
u
′
(
t
)
=
2
(
t
−
1
)
3
(
−
3
t
+
23
)
(
3
t
+
7
)
7
\displaystyle u'\left(t\right)=\frac{2\left(t-1\right)^3\left(-3t+23\right)}{\left(3t+7\right)^7}
u
′
(
t
)
=
(
3
t
+
7
)
7
2
(
t
−
1
)
3
(
−
3
t
+
23
)
the derivative of
u
(
t
)
=
(
t
−
1
)
4
(
3
t
+
7
)
6
\displaystyle u\left(t\right)=\frac{\left(t-1\right)^4}{\left(3t+7\right)^6}
u
(
t
)
=
(
3
t
+
7
)
6
(
t
−
1
)
4
?
The quotient rule
Find the derivative of the following function:
g
(
x
)
=
x
+
1
x
+
1
\displaystyle g(x)=\frac{\sqrt{x+1}}{\sqrt{x}+1}
g
(
x
)
=
x
+
1
x
+
1
The quotient rule
Is
u
′
(
t
)
=
2
(
t
−
1
)
3
(
−
3
t
+
23
)
(
3
t
+
7
)
7
\displaystyle u'\left(t\right)=\frac{2\left(t-1\right)^3\left(-3t+23\right)}{\left(3t+7\right)^7}
u
′
(
t
)
=
(
3
t
+
7
)
7
2
(
t
−
1
)
3
(
−
3
t
+
23
)
the derivative of
u
(
t
)
=
(
t
−
1
)
4
(
3
t
+
7
)
6
\displaystyle u\left(t\right)=\frac{\left(t-1\right)^4}{\left(3t+7\right)^6}
u
(
t
)
=
(
3
t
+
7
)
6
(
t
−
1
)
4
?
The quotient rule
Given the function
f
(
x
)
=
g
(
x
)
h
(
x
)
−
e
x
cos
x
f\left(x\right)=\dfrac{g\left(x\right)}{h\left(x\right)}-e^x\cos x
f
(
x
)
=
h
(
x
)
g
(
x
)
−
e
x
cos
x
, and given
g
(
π
2
)
=
2
g\left(\dfrac{\pi}{2}\right)=2
g
(
2
π
)
=
2
,
g
′
(
π
2
)
=
3
g'\left(\dfrac{\pi}{2}\right)=3
g
′
(
2
π
)
=
3
,
h
(
π
2
)
=
1
h\left(\dfrac{\pi}{2}\right)=1
h
(
2
π
)
=
1
, and
h
′
(
π
2
)
=
2
h'\left(\dfrac{\pi}{2}\right)=2
h
′
(
2
π
)
=
2
, find
f
′
(
π
2
)
f'\left(\dfrac{\pi}{2}\right)
f
′
(
2
π
)
.
Derivatives: Exponential and Logarithmic Functions
Find
f
′
(
x
)
f'(x)
f
′
(
x
)
if
f
(
x
)
=
ln
x
x
e
x
\displaystyle f\left(x\right)=\frac{\ln x}{xe^{x}}
f
(
x
)
=
x
e
x
ln
x
. Simplify.
Quotient and Product
Find the derivative of
g
(
t
)
=
(
1
+
2
t
7
t
)
(
t
2
−
1
)
\displaystyle g(t)=\left(\frac{1+2t}{7t}\right)(t^{2}-1)
g
(
t
)
=
(
7
t
1
+
2
t
)
(
t
2
−
1
)
Practice: Quotient Rule
Q.
\textbf{Q.}
Q.
Find the derivative of
f
(
x
)
=
x
2
+
3
x
4
/
3
+
1
x
2
+
1
\displaystyle f(x)=\frac{x^2+3x^{4/3}+1}{x^2+1}
f
(
x
)
=
x
2
+
1
x
2
+
3
x
4/3
+
1
Find the derivative of the following function:
g
(
x
)
=
x
+
1
x
+
1
\displaystyle g(x)=\frac{\sqrt{x+1}}{\sqrt{x}+1}
g
(
x
)
=
x
+
1
x
+
1
The Quotient Rule
Find the derivative of
g
(
x
)
=
x
2
−
5
2
x
+
1
g(x) = \frac{x^2 - 5}{2x + 1}
g
(
x
)
=
2
x
+
1
x
2
−
5
Derivatives
Find the derivative of the following functions at the given values
a)
f
(
x
)
=
arcsin
(
x
2
)
x
3
−
1
f\left(x\right)=\frac{\arcsin\left(x^2\right)}{x^3-1}
f
(
x
)
=
x
3
−
1
a
r
c
s
i
n
(
x
2
)
at
x
=
0
x=0
x
=
0
b)
g
(
x
)
=
(
cos
x
)
e
x
g\left(x\right)=\left(\cos x\right)^{e^x}
g
(
x
)
=
(
cos
x
)
e
x
at
x
=
0
x=0
x
=
0
The Quotient Rule
Compute the derivative of
f
(
x
)
=
x
2
+
3
x
4
/
3
+
1
x
2
+
1
f(x)=\frac{x^2+3x^{4/3}+1}{x^2+1}
f
(
x
)
=
x
2
+
1
x
2
+
3
x
4/3
+
1
Practice: Product Rule*
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
1
+
x
g
(
x
)
x
2
f\left(x\right)=\frac{1+x\ g\left(x\right)}{x^2}
f
(
x
)
=
x
2
1
+
x
g
(
x
)
,
f
′
(
1
)
=
2
f'\left(1\right)=2
f
′
(
1
)
=
2
,
g
(
1
)
=
1
g\left(1\right)=1
g
(
1
)
=
1
, what is
g
′
(
1
)
g'\left(1\right)
g
′
(
1
)
?
The quotient rule
Given the function
f
(
x
)
=
g
(
x
)
h
(
x
)
−
e
x
cos
x
f\left(x\right)=\dfrac{g\left(x\right)}{h\left(x\right)}-e^x\cos x
f
(
x
)
=
h
(
x
)
g
(
x
)
−
e
x
cos
x
, and given
g
(
π
2
)
=
2
g\left(\dfrac{\pi}{2}\right)=2
g
(
2
π
)
=
2
,
g
′
(
π
2
)
=
3
g'\left(\dfrac{\pi}{2}\right)=3
g
′
(
2
π
)
=
3
,
h
(
π
2
)
=
1
h\left(\dfrac{\pi}{2}\right)=1
h
(
2
π
)
=
1
, and
h
′
(
π
2
)
=
2
h'\left(\dfrac{\pi}{2}\right)=2
h
′
(
2
π
)
=
2
, find
f
′
(
π
2
)
f'\left(\dfrac{\pi}{2}\right)
f
′
(
2
π
)
.
Derivatives: Exponential and Logarithmic Functions
Find
f
′
(
x
)
f'(x)
f
′
(
x
)
if
f
(
x
)
=
ln
x
x
e
x
\displaystyle f\left(x\right)=\frac{\ln x}{xe^{x}}
f
(
x
)
=
x
e
x
ln
x
. Simplify.
The quotient rule
Is
u
′
(
t
)
=
2
(
t
−
1
)
3
(
−
3
t
+
23
)
(
3
t
+
7
)
7
\displaystyle u'\left(t\right)=\frac{2\left(t-1\right)^3\left(-3t+23\right)}{\left(3t+7\right)^7}
u
′
(
t
)
=
(
3
t
+
7
)
7
2
(
t
−
1
)
3
(
−
3
t
+
23
)
the derivative of
u
(
t
)
=
(
t
−
1
)
4
(
3
t
+
7
)
6
\displaystyle u\left(t\right)=\frac{\left(t-1\right)^4}{\left(3t+7\right)^6}
u
(
t
)
=
(
3
t
+
7
)
6
(
t
−
1
)
4
?
Quotient and Product
Find the derivative of
g
(
t
)
=
(
1
+
2
t
7
t
)
(
t
2
−
1
)
\displaystyle g(t)=\left(\frac{1+2t}{7t}\right)(t^{2}-1)
g
(
t
)
=
(
7
t
1
+
2
t
)
(
t
2
−
1
)
Practice: Quotient Rule
Find the derivative of
f
(
x
)
=
x
2
+
3
x
4
/
3
+
1
x
2
+
1
\displaystyle f(x)=\frac{x^2+3x^{4/3}+1}{x^2+1}
f
(
x
)
=
x
2
+
1
x
2
+
3
x
4/3
+
1
The Quotient Rule
If
h
(
x
)
=
f
(
x
)
−
1
x
+
2
h\left(x\right)=\frac{f\left(x\right)-1}{x+2}
h
(
x
)
=
x
+
2
f
(
x
)
−
1
,
f
(
3
)
=
2
and
f
′
(
3
)
=
−
6
f\left(3\right)=2\ \text{and}\ f'\left(3\right)=-6
f
(
3
)
=
2
and
f
′
(
3
)
=
−
6
, calculate the value of
h
′
(
3
)
h'\left(3\right)
h
′
(
3
)
The Quotient Rule
Is
u
′
(
t
)
=
2
(
t
−
1
)
3
(
−
3
t
+
23
)
(
3
t
+
7
)
7
u'\left(t\right)=\frac{2\left(t-1\right)^3\left(-3t+23\right)}{\left(3t+7\right)^7}
u
′
(
t
)
=
(
3
t
+
7
)
7
2
(
t
−
1
)
3
(
−
3
t
+
23
)
the derivative of
u
(
t
)
=
(
t
−
1
)
4
(
3
t
+
7
)
6
u\left(t\right)=\frac{\left(t-1\right)^4}{\left(3t+7\right)^6}
u
(
t
)
=
(
3
t
+
7
)
6
(
t
−
1
)
4
?
Find the derivative with respect to
x
x
x
of the function
x
2030
3
+
x
2030
\displaystyle \frac{x^{2030}}{3+x^{2030}}
3
+
x
2030
x
2030
.
Compute the derivative of
f
(
x
)
=
x
2
+
3
x
4
/
3
+
1
x
2
+
1
f(x) =\dfrac{x^2 + 3x^{4/3} + 1}{x^2 + 1}
f
(
x
)
=
x
2
+
1
x
2
+
3
x
4/3
+
1
.
The Quotient Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
1
+
x
g
(
x
)
x
2
f\left(x\right)=\frac{1+x\ g\left(x\right)}{x^2}
f
(
x
)
=
x
2
1
+
x
g
(
x
)
,
f
′
(
1
)
=
2
f'\left(1\right)=2
f
′
(
1
)
=
2
,
g
(
1
)
=
1
g\left(1\right)=1
g
(
1
)
=
1
, what is
g
′
(
1
)
g'\left(1\right)
g
′
(
1
)
?
Find the derivative of the following function:
g
(
x
)
=
x
+
1
x
+
1
\displaystyle g(x)=\frac{\sqrt{x+1}}{\sqrt{x}+1}
g
(
x
)
=
x
+
1
x
+
1
The quotient rule
Find the derivative of the following function:
g
(
x
)
=
x
+
1
x
+
1
\displaystyle g(x)=\frac{\sqrt{x+1}}{\sqrt{x}+1}
g
(
x
)
=
x
+
1
x
+
1