High School
SAT
SAT Elite 1500
SAT Tutoring
ACT
ACT Elite 33
ACT Tutoring
University
MCAT
MCAT Elite 515
Med-School Admissions
Pre-Med Tutoring
Pre-Med Plus
LSAT
LSAT Elite 170
LSAT Self-Paced
LSAT Tutoring
DAT
DAT Elite
DAT Tutoring
Log in
Get Started for Free
Find the derivative of the following function. f(x)=√arcsinx
Related Topics
Wize University Calculus 1 Textbook > Derivatives
The Chain Rule
3 Activities
Wize University Calculus 1 Textbook > Derivatives
Derivatives of Inverse Trig Functions
3 Activities
Find the derivative of the following function.
f
(
x
)
=
arcsinx
\displaystyle f(x)=\sqrt{\text{arcsin{x}}}
f
(
x
)
=
arcsin
x
1
2
[
arcsin
(
x
)
]
1
/
2
⋅
1
1
+
x
2
\dfrac{1}{2}[\arcsin(x)]^{1/2}\cdot\dfrac{1}{\sqrt{1+x^2}}
2
1
[
arcsin
(
x
)
]
1/2
⋅
1
+
x
2
1
1
2
[
arcsin
(
x
)
]
−
1
/
2
⋅
1
1
−
x
2
\frac{1}{2}[\arcsin(x)]^{-1/2}\cdot\frac{1}{\sqrt{1-x^2}}
2
1
[
arcsin
(
x
)
]
−
1/2
⋅
1
−
x
2
1
[
arcsin
(
x
)
]
1
/
2
⋅
1
1
−
x
2
[\arcsin(x)]^{1/2}\cdot\frac{1}{\sqrt{1-x^2}}
[
arcsin
(
x
)
]
1/2
⋅
1
−
x
2
1
1
2
[
arcsin
(
x
)
]
⋅
1
1
−
x
2
\dfrac{1}{2}[\arcsin(x)]\cdot\dfrac{1}{\sqrt{1-x^2}}
2
1
[
arcsin
(
x
)]
⋅
1
−
x
2
1
I don't know
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 Quotient and Chain Rules
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
.
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
)
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 Derivatives of Inverse Trig Functions Questions:
Inverse trigonometric derivative
Find the derivative of
g
(
x
)
=
(
cos
−
1
x
)
(
sin
−
1
x
)
g(x)=\left(\cos^{-1}x\right)\left(\sin^{-1}x\right)
g
(
x
)
=
(
cos
−
1
x
)
(
sin
−
1
x
)
.
Derivatives: Inverse Trigonometric Functions
Given that
f
(
x
)
=
2
−
arcsin
(
x
)
f\left(x\right)=2-\arcsin\left(x\right)
f
(
x
)
=
2
−
arcsin
(
x
)
, find
f
′
(
0
)
f'\left(0\right)
f
′
(
0
)
Inverse Trigonometric Functions
Compute the derivative of
f
(
x
)
=
arcsin
(
x
+
2
)
f\left(x\right)=\arcsin\left(x+2\right)
f
(
x
)
=
arcsin
(
x
+
2
)
Derivatives: Inverse Trigonometric Functions
Given that
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
f\left(x\right)=\left(2-\arcsin\left(x^2\right)\right)^3
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
, find
f
′
(
0
)
f'\left(0\right)
f
′
(
0
)
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
)
′
Inverse trigonometric derivative
Find the derivative of
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
g(x)=\cos^{-1}\left(\sin^{-1}x\right)
g
(
x
)
=
cos
−
1
(
sin
−
1
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
)
)
.
Inverse trigonometric derivative
Find the derivative of
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
g(x)=\cos^{-1}\left(\sin^{-1}x\right)
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
.
Inverse Trigonometric Functions
Compute the derivative of
f
(
x
)
=
arcsin
(
x
+
2
x
−
1
)
\displaystyle f(x)=\text{arcsin}\left(\frac{x+2}{x-1}\right)
f
(
x
)
=
arcsin
(
x
−
1
x
+
2
)
Derivatives: Inverse Trigonometric Functions
Given that
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
f\left(x\right)=\left(2-\arcsin\left(x^2\right)\right)^3
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
, find
f
′
(
0
)
f'\left(0\right)
f
′
(
0
)
.
Evaluating a derivative at a point
Evaluate
(
cos
−
1
(
x
2
)
)
′
∣
0
\left( \cos^{-1}\left(x^2\right) \right)'\rvert_{0}
(
cos
−
1
(
x
2
)
)
′
∣
0
Inverse Trig Derivative
Find the equation of the line tangent to
f
(
x
)
=
sin
−
1
x
1
+
x
f(x)=\frac{\sin^{-1}x}{1+x}
f
(
x
)
=
1
+
x
sin
−
1
x
at
x
=
1
/
2
x=1/2
x
=
1/2
.
Inverse Trigonometric Functions
Compute the derivative of
f
(
x
)
=
arcsin
(
x
+
2
x
−
1
)
\displaystyle f(x)=\text{arcsin}\left(\frac{x+2}{x-1}\right)
f
(
x
)
=
arcsin
(
x
−
1
x
+
2
)
Derivatives: Inverse Trigonometric Functions
Practice: Finding a Derivative
Given that
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
f\left(x\right)=\left(2-\arcsin\left(x^2\right)\right)^3
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
, find
f
′
(
0
)
f'\left(0\right)
f
′
(
0
)
.
Evaluating a derivative at a point
Evaluate
(
cos
−
1
(
x
2
)
)
′
∣
0
\left( \cos^{-1}\left(x^2\right) \right)'\rvert_{0}
(
cos
−
1
(
x
2
)
)
′
∣
0
Inverse Trig Derivative
Find the equation of the line tangent to
f
(
x
)
=
sin
−
1
x
1
+
x
f(x)=\frac{\sin^{-1}x}{1+x}
f
(
x
)
=
1
+
x
sin
−
1
x
at
x
=
1
/
2
x=1/2
x
=
1/2
.
Inverse trigonometric derivative
Find the derivative of
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
g(x)=\cos^{-1}\left(\sin^{-1}x\right)
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
.
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
)
)
.
Evaluating a derivative at a point
Evaluate
(
cos
−
1
(
x
2
)
)
′
∣
0
\left( \cos^{-1}\left(x^2\right) \right)'\rvert_{0}
(
cos
−
1
(
x
2
)
)
′
∣
0
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
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
)
′
Derivatives: Inverse Trigonometric Functions
Find
[
tan
−
1
(
x
3
)
]
′
[\tan^{-1}(\sqrt[3]x)]'
[
tan
−
1
(
3
x
)
]
′
Derivatives: Inverse Trigonometric Functions
Show that
d
d
x
sin
−
1
x
=
1
1
−
x
2
\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}
d
x
d
sin
−
1
x
=
1
−
x
2
1
Practice: Derivative of Inverse Trig
Q:
\textbf{Q:}
Q:
Show that
d
d
x
cos
−
1
x
=
−
1
1
−
x
2
.
\displaystyle\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}.
d
x
d
cos
−
1
x
=
−
1
−
x
2
1
.
Practice: Tangent Line with Inverse Trig
Q:
\textbf{Q:}
Q:
Find the equation of the tangent line to the graph of
f
(
x
)
=
cot
−
1
(
x
2
)
\displaystyle f(x)=\cot^{-1}(x^{2})
f
(
x
)
=
cot
−
1
(
x
2
)
at the point
x
=
2
x=2
x
=
2
.
Inverse Trig with Log
Compute the derivative of
f
(
x
)
=
arctan
(
log
10
x
)
\displaystyle f(x)=\text{arctan}\left(\text{log}_{10}x\right)
f
(
x
)
=
arctan
(
log
10
x
)
Inverse Trigonometric Functions
Compute the derivative of
f
(
x
)
=
arcsin
(
x
+
2
x
−
1
)
\displaystyle f(x)=\text{arcsin}\left(\frac{x+2}{x-1}\right)
f
(
x
)
=
arcsin
(
x
−
1
x
+
2
)
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
Derivatives: Exponential and Trigonometric Functions
Find the derivative of
f
(
x
)
=
5
x
6
−
sec
−
1
(
e
2
x
)
+
cos
−
1
x
f(x)=5x^6-\sec^{-1}(e^{2x})+\cos^{-1}x
f
(
x
)
=
5
x
6
−
sec
−
1
(
e
2
x
)
+
cos
−
1
x
Inverse Trig Derivative
Find the equation of the line tangent to
f
(
x
)
=
sin
−
1
x
1
+
x
f(x)=\frac{\sin^{-1}x}{1+x}
f
(
x
)
=
1
+
x
sin
−
1
x
at
x
=
1
/
2
x=1/2
x
=
1/2
.
Derivatives: Inverse Trigonometric Functions
Find
f
′
(
x
)
f'(x)
f
′
(
x
)
if
f
(
x
)
=
arctan
x
f(x)=\arctan \sqrt{x}
f
(
x
)
=
arctan
x
Differentiate the following
y
=
tan
−
1
x
2
−
1
y=\tan^{-1}\sqrt{x^2-1}
y
=
tan
−
1
x
2
−
1
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
)
′
Derivatives: Inverse Trigonometric Functions
Find
[
tan
−
1
(
x
3
)
]
′
[\tan^{-1}(\sqrt[3]x)]'
[
tan
−
1
(
3
x
)
]
′
Derivatives: Inverse Trigonometric Functions
Show that
d
d
x
sin
−
1
x
=
1
1
−
x
2
\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}
d
x
d
sin
−
1
x
=
1
−
x
2
1
Practice: Derivative of Inverse Trig
Q:
\textbf{Q:}
Q:
Show that
d
d
x
cos
−
1
x
=
−
1
1
−
x
2
.
\displaystyle\frac{d}{dx}\cos^{-1}x=-\frac{1}{\sqrt{1-x^2}}.
d
x
d
cos
−
1
x
=
−
1
−
x
2
1
.
Practice: Tangent Line with Inverse Trig
Q:
\textbf{Q:}
Q:
Find the equation of the tangent line to the graph of
f
(
x
)
=
cot
−
1
(
x
2
)
\displaystyle f(x)=\cot^{-1}(x^{2})
f
(
x
)
=
cot
−
1
(
x
2
)
at the point
x
=
2
x=2
x
=
2
.
Inverse Trig with Log
Compute the derivative of
f
(
x
)
=
arctan
(
log
10
x
)
\displaystyle f(x)=\text{arctan}\left(\text{log}_{10}x\right)
f
(
x
)
=
arctan
(
log
10
x
)
Inverse Trigonometric Functions
Compute the derivative of
f
(
x
)
=
arcsin
(
x
+
2
x
−
1
)
\displaystyle f(x)=\text{arcsin}\left(\frac{x+2}{x-1}\right)
f
(
x
)
=
arcsin
(
x
−
1
x
+
2
)
Derivatives: Exponential Functions, Inverse Trigonometric Functions
Compute the derivative of
f
(
x
)
=
sec
(
3
2
x
)
f(x)=\sec(3^{2x})
f
(
x
)
=
sec
(
3
2
x
)
.
Derivatives: Logarithmic and Inverse Trigonometric Functions
Evaluate
d
d
x
(
sin
−
1
x
sin
x
log
3
x
)
\displaystyle \frac{\text{d}}{\text{d}x}\left( \frac{\sin^{-1}x\sin x}{\log_3x}\right)
d
x
d
(
lo
g
3
x
sin
−
1
x
sin
x
)
.
Find the derivative of
f
(
x
)
=
arcsin
(
e
x
+
1
)
f(x)=\arcsin(e^{x+1})
f
(
x
)
=
arcsin
(
e
x
+
1
)
.
Inverse trigonometric derivative
Find the derivative of
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
g(x)=\cos^{-1}\left(\sin^{-1}x\right)
g
(
x
)
=
cos
−
1
(
sin
−
1
x
)
.
Derivatives: Inverse Trigonometric Functions
Find the derivative of
f
(
x
)
=
(
arccos
x
)
2
sec
x
f\left(x\right)=\frac{\left(\arccos x\right)^2}{\sec x}
f
(
x
)
=
s
e
c
x
(
a
r
c
c
o
s
x
)
2
.
Evaluating a derivative at a point
Evaluate
(
cos
−
1
(
x
2
)
)
′
∣
0
\left( \cos^{-1}\left(x^2\right) \right)'\rvert_{0}
(
cos
−
1
(
x
2
)
)
′
∣
0
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
)
Derivatives: Inverse Trigonometric Functions
Practice: Finding a Derivative
Given that
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
f\left(x\right)=\left(2-\arcsin\left(x^2\right)\right)^3
f
(
x
)
=
(
2
−
arcsin
(
x
2
)
)
3
, find
f
′
(
0
)
f'\left(0\right)
f
′
(
0
)
.
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
=
Linear Approximation and Trigonometric Derivative
a) Show that for
f
(
x
)
=
arcsin
(
x
)
f(x) = \arcsin(x)
f
(
x
)
=
arcsin
(
x
)
, that
f
′
(
x
)
=
1
1
−
x
2
f'(x) = \frac{1}{\sqrt{1 - x^2}}
f
′
(
x
)
=
1
−
x
2
1
.
b) Use linear approximation at a suitable close value to estimate
arcsin
(
0.1
)
\arcsin(0.1)
arcsin
(
0.1
)
. Your solution may be left in terms of fractions.
Derivatives: Inverse Trigonometric Functions
Find the derivative of the following function.
f
(
x
)
=
1
arcsin
(
x
)
+
arcsin
(
1
x
)
\displaystyle f(x)=\frac{1}{\text{arcsin}(x)}+\text{arcsin}\left(\frac{1}{x}\right)
f
(
x
)
=
arcsin
(
x
)
1
+
arcsin
(
x
1
)
Inverse Trigonometric Functions
Calculate the derivative of the following functions.
f
(
x
)
=
arcsin
(
e
x
+
1
)
\displaystyle f(x)=\text{arcsin}(e^{x+1})
f
(
x
)
=
arcsin
(
e
x
+
1
)
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
))