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The product rule
Related Topics
Wize University Calculus 1 Textbook > Derivatives
The Product Rule
4 Activities
Wize University Calculus 1 Textbook > Derivatives
Differentiation Laws
1 Activity
Find the derivative of
f
(
x
)
=
x
3
e
x
f(x) = x^3 e^x
f
(
x
)
=
x
3
e
x
x
2
e
x
(
x
+
3
)
x^2 e^x (x+3)
x
2
e
x
(
x
+
3
)
e
x
(
3
x
2
+
1
)
e^x (3x^2 + 1)
e
x
(
3
x
2
+
1
)
3
x
2
e
x
+
x
3
e
x
−
1
3x^2 e^x + x^3 e^{x-1}
3
x
2
e
x
+
x
3
e
x
−
1
3
x
2
e
x
−
1
+
x
3
e
x
3x^2 e^{x-1} + x^3 e^x
3
x
2
e
x
−
1
+
x
3
e
x
e
x
(
x
+
3
)
e^x (x + 3)
e
x
(
x
+
3
)
x
e
x
(
3
x
+
1
)
x e^x (3x + 1)
x
e
x
(
3
x
+
1
)
I don't know
Check Submission
More The Product Rule Questions:
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Q:
\textbf{Q:}
Q:
Differentiate
f
(
x
)
=
(
x
3
/
2
)
(
2
x
4
−
5
x
2
−
x
−
1
)
(
x
e
−
π
3.5
)
\displaystyle f(x)=(x^{3/2})\left(2x^{4}-5x^{2}-x^{-1}\right)(x^{e}-\pi^{3.5})
f
(
x
)
=
(
x
3/2
)
(
2
x
4
−
5
x
2
−
x
−
1
)
(
x
e
−
π
3.5
)
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
)
The graph of w(x) is given below.
Practice: Product Rule*
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
Product Rule
Differentiate
f
(
x
)
=
(
x
3
/
2
)
(
2
x
4
−
5
x
2
−
x
−
1
)
(
x
e
−
π
3.5
)
\displaystyle f(x)=(x^{3/2})\left(2x^{4}-5x^{2}-x^{-1}\right)(x^{e}-\pi^{3.5})
f
(
x
)
=
(
x
3/2
)
(
2
x
4
−
5
x
2
−
x
−
1
)
(
x
e
−
π
3.5
)
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
)
Find the derivative of
f
(
x
)
=
(
x
3
/
2
+
2
x
)
(
x
7
/
4
+
x
)
\displaystyle f(x)=\left(x^{3/2}+\frac{2}{x}\right)\left(x^{7/4}+x\right)
f
(
x
)
=
(
x
3/2
+
x
2
)
(
x
7/4
+
x
)
.
Product Rule
Suppose
g
(
x
)
g\left(x\right)
g
(
x
)
is differentiable,
f
(
x
)
=
x
2
g
(
x
)
f\left(x\right)=x^2g\left(x\right)
f
(
x
)
=
x
2
g
(
x
)
. Given that
f
′
(
3
)
=
30
f'\left(3\right)=30
f
′
(
3
)
=
30
,
f
′
′
(
3
)
=
19
f''\left(3\right)=19
f
′′
(
3
)
=
19
,
g
(
3
)
=
2
g\left(3\right)=2
g
(
3
)
=
2
, what is
g
′
(
3
)
g'\left(3\right)
g
′
(
3
)
and
g
′
′
(
3
)
g''\left(3\right)
g
′′
(
3
)
?
More Differentiation Laws Questions:
Derivative of Polynomials: Product Rule
Find the derivative of
f
(
x
)
=
(
x
3
/
2
+
2
x
)
(
x
7
/
4
+
x
)
f(x)=\left(x^{3/2}+\frac{2}{x}\right)\left(x^{7/4}+x\right)
f
(
x
)
=
(
x
3/2
+
x
2
)
(
x
7/4
+
x
)
Differentiation laws
What is the derivative of the function
f
(
x
)
=
x
3
+
1
x
2
f(x)=\frac{x^3+1}{x^2}
f
(
x
)
=
x
2
x
3
+
1
?
Differentiation: Piecewise Differentiable Function
Find
A
A
A
and
B
B
B
for which
f
(
x
)
f\left(x\right)
f
(
x
)
is differentiable everywhere. Answers are in the form
(
A
,
B
)
\left(A,B\right)
(
A
,
B
)
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
f(x)=\begin{cases} xe^{x^2+1}, \text{ if } x\geq 1 \\ Ax+B, \text{ if } x < 1 \end{cases}
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
Differentiation: Piecewise Differentiable Function
Find
A
A
A
and
B
B
B
for which
f
(
x
)
f\left(x\right)
f
(
x
)
is differentiable everywhere.
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
f(x)=\begin{cases} xe^{x^2+1}, \text{ if } x\geq 1 \\ Ax+B, \text{ if } x < 1 \end{cases}
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
Derivative of Polynomials: Product Rule
Find the derivative of
f
(
x
)
=
(
x
3
/
2
+
2
x
)
(
x
7
/
4
+
x
)
f(x)=\left(x^{3/2}+\frac{2}{x}\right)\left(x^{7/4}+x\right)
f
(
x
)
=
(
x
3/2
+
x
2
)
(
x
7/4
+
x
)
Differentiation: Piecewise Differentiable Function
Find
A
A
A
and
B
B
B
for which
f
(
x
)
f\left(x\right)
f
(
x
)
is differentiable everywhere.
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
f(x)=\begin{cases} xe^{x^2+1}, \text{ if } x\geq 1 \\ Ax+B, \text{ if } x < 1 \end{cases}
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
Differentiation: Piecewise Differentiable Function
Find
A
A
A
and
B
B
B
for which
f
(
x
)
f\left(x\right)
f
(
x
)
is differentiable everywhere.
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
f(x)=\begin{cases} xe^{x^2+1}, \text{ if } x\geq 1 \\ Ax+B, \text{ if } x < 1 \end{cases}
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
Differentiation: Piecewise Differentiable Function
Find
A
A
A
and
B
B
B
for which
f
(
x
)
f\left(x\right)
f
(
x
)
is differentiable everywhere.
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
f(x)=\begin{cases} xe^{x^2+1}, \text{ if } x\geq 1 \\ Ax+B, \text{ if } x < 1 \end{cases}
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
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
.
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
.
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
Differentiation: Piecewise Differentiable Function
Find
A
A
A
and
B
B
B
for which
f
(
x
)
f\left(x\right)
f
(
x
)
is differentiable everywhere.
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
f(x)=\begin{cases} xe^{x^2+1}, \text{ if } x\geq 1 \\ Ax+B, \text{ if } x < 1 \end{cases}
f
(
x
)
=
{
x
e
x
2
+
1
,
if
x
≥
1
A
x
+
B
,
if
x
<
1
Derivative of Polynomials: Product Rule
Find the derivative of
f
(
x
)
=
(
x
3
/
2
+
2
x
)
(
x
7
/
4
+
x
)
f(x)=\left(x^{3/2}+\frac{2}{x}\right)\left(x^{7/4}+x\right)
f
(
x
)
=
(
x
3/2
+
x
2
)
(
x
7/4
+
x
)
Differentiation laws
What is the derivative of the function
f
(
x
)
=
x
3
+
1
x
2
f(x)=\frac{x^3+1}{x^2}
f
(
x
)
=
x
2
x
3
+
1
?
Is
f
(
x
)
f(x)
f
(
x
)
differentiable at
x
=
1
x=1
x
=
1
? If so, find
f
′
(
1
)
f'(1)
f
′
(
1
)
.
f
(
x
)
=
{
x
+
1
if
x
<
1
1
2
x
2
+
3
2
if
x
≥
1
f(x) = \begin{cases} x+1 & \text{if } x < 1 \\ \frac{1}{2}x^2 + \frac{3}{2} & \text{if } x \geq 1 \end{cases}
f
(
x
)
=
{
x
+
1
2
1
x
2
+
2
3
if
x
<
1
if
x
≥
1
Differential Laws: nth Derivatives
If
y
=
(
10
e
+
1
)
10
y=\left(10e+1\right)^{10}
y
=
(
10
e
+
1
)
10
, find the 9
th
derivative of
y
y
y
.
If
f
(
x
)
f(x)
f
(
x
)
is differentiable everywhere, find
A
A
A
and
B
B
B
.
f
(
x
)
=
{
x
2
+
1
if
x
≥
0
A
x
+
B
if
x
<
0
f(x)=\left\{ \begin{array}{ll} \displaystyle x^2+1\quad\quad\,\,\,\,\,\text{if}\,x\geq 0 \\ Ax+B\quad\,\,\,\,\,\,\,\text{if}\,\, x<0\\ \end{array} \right.
f
(
x
)
=
{
x
2
+
1
if
x
≥
0
A
x
+
B
if
x
<
0