How do you find f'(x) and f''(x) given f(x)=x3+ex?
Answers
Answer:
Use the product rule to take the first derivative, than again to take the second. This yields
f
'
(
x
)
=
4
x
3
e
x
+
x
4
e
x
and
f
'
'
(
x
)
=
x
4
e
x
+
8
x
3
e
x
+
12
x
2
e
x
.
Explanation:
By the product rule, for a function
f
g
, the derivative is given by
f
'
g
+
f
g
'
or
f
g
'
+
f
'
g
. The order does not matter because addition is commutative.
To find the first derivative of the function,
f
'
(
x
)
, first take the derivative of
x
4
and multiply this by
e
x
, which we have left unchanged. This gives:
4
x
3
⋅
e
x
Now, take the derivative of
e
x
(which is still
e
x
by properties of exponential functions) and multiply this by
x
4
, which we have left unchanged. This gives:
e
x
⋅
x
4
Adding these two halves together, we get a final answer of:
f
'
(
x
)
=
4
x
3
e
x
+
x
4
e
x
Finding the second derivative follows a similar process. Our starting function is now
f
'
(
x
)
. We will use the product rule to take the derivative of each "half" that we found above for
f
'
(
x
)
, and add those halves together (as specified in the function).
First, we take the derivative of
4
x
3
e
x
. Using the product rule, we take the derivative of
4
x
3
and multiply this by
e
x
, which we leave unchanged, then add this to the derivative of
e
x
, which we multiply by
4
x
3
. This gives:
12
x
2
e
x
+
4
x
3
e
x
Now, we take the derivative of
x
4
e
x
following the same process. We get:
4
x
3
e
x
+
x
4
e
x
Now, we add these two halves together to get:
f
'
'
(
x
)
=
12
x
2
e
x
+
4
x
3
e
x
+
4
x
3
e
x
+
x
4
e
x
This simplifies to:
f
'
'
(
x
)
=
x
4
e
x
+
8
x
3
e
x
+
12
x
2
e
x
or further to:
f
'
'
(
x
)
=
e
x
(
x
4
+
8
x
3
+
12
x
2
)
Hope that helps!