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Higher Order Derivatives: The Power Rule
Related Topics
Wize University Calculus 1 Textbook > Derivatives
The Power Rule (Derivative of Polynomials)
3 Activities
Wize University Calculus 1 Textbook > Derivatives
Higher Order Derivatives
4 Activities
If
f
(
x
)
=
x
10
f\left(x\right)=x^{10}
f
(
x
)
=
x
10
find
f
(
9
)
(
x
)
f^{\left(9\right)}\left(x\right)
f
(
9
)
(
x
)
.
(i.e. the 9th derivative)
10
!
x
10!\ x
10
!
x
2
10
x
2^{10}\ x
2
10
x
9
!
x
2
9!\ x^2
9
!
x
2
2
9
x
2
2^9\ x^2
2
9
x
2
0
0
0
I don't know
Check Submission
More The Power Rule (Derivative of Polynomials) 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
)
The Power Rule: Derivatives of Polynomials
Find the derivative of
f
(
x
)
=
x
3
x
f\left(x\right)=\dfrac{x^3}{\sqrt{x}}
f
(
x
)
=
x
x
3
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
)
Higher Order Derivatives: The Power Rule
If
f
(
x
)
=
x
10
f\left(x\right)=x^{10}
f
(
x
)
=
x
10
find
f
(
9
)
(
x
)
f^{\left(9\right)}\left(x\right)
f
(
9
)
(
x
)
.
(i.e. the 9th derivative)
Find the derivative of
f
(
x
)
=
x
3
x
f\left(x\right)=\frac{x^3}{\sqrt{x}}
f
(
x
)
=
x
x
3
Derivative of Polynomials
Suppose
f
is a function which satisfies
f
′
(
0
)
=
π
and
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
f'(0)=\pi \text{ and }f(x+y)=f(x)+f(y)
f
′
(
0
)
=
π
and
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
for all
x
and
y
. True or false:
f
(
x
)
=
π
x
f(x)=\pi x
f
(
x
)
=
π
x
satisfies this function.
The Power Rule: Derivatives of Polynomials
Find the derivative of
f
(
x
)
=
x
3
x
f\left(x\right)=\dfrac{x^3}{\sqrt{x}}
f
(
x
)
=
x
x
3
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
)
If 𝑓(𝑥) = 𝑥
10
, find 𝑓
(9)
(𝑥).
(i.e. the 9th derivative)
Find the derivative of
f
(
x
)
=
x
3
x
f\left(x\right)=\frac{x^3}{\sqrt{x}}
f
(
x
)
=
x
x
3
More Higher Order Derivatives Questions:
Higher Order Derivatives
Find the second derivative of
f
(
x
)
=
x
2
+
16
x
f(x) = x^2+16\sqrt{x}
f
(
x
)
=
x
2
+
16
x
.
Higher order derivatives
Consider the function
f
(
x
)
=
3
x
−
4
x
f(x)=\dfrac{3x-4}{x}
f
(
x
)
=
x
3
x
−
4
. Find the second derivative
f
′
′
(
x
)
,
at
x
=
1
f''(x),\text{ at }x=1
f
′′
(
x
)
,
at
x
=
1
.
Higher order derivatives
Consider the function
f
(
x
)
=
3
x
−
4
x
f(x)=\dfrac{3x-4}{x}
f
(
x
)
=
x
3
x
−
4
. Find the second derivative
f
′
′
(
x
)
,
at
x
=
1
f''(x),\text{ at }x=1
f
′′
(
x
)
,
at
x
=
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
Higher Order Derivatives
If
f
(
x
)
=
sin
(
2
x
)
f\left(x\right)=\sin\left(2x\right)
f
(
x
)
=
sin
(
2
x
)
, find
f
(
21
)
(
π
2
)
f^{\left(21\right)}\left(\frac{\pi}{2}\right)
f
(
21
)
(
2
π
)
.
(i.e. find the 21st derivative at the point
π
2
\frac{\pi}{2}
2
π
)
Higher Order Derivatives
If
f
(
x
)
=
sin
(
2
x
)
f\left(x\right)=\sin\left(2x\right)
f
(
x
)
=
sin
(
2
x
)
, find
f
(
21
)
(
π
2
)
f^{\left(21\right)}\left(\frac{\pi}{2}\right)
f
(
21
)
(
2
π
)
.
(i.e. find the 21st derivative at the point
π
2
\frac{\pi}{2}
2
π
)
Higher order derivatives
Consider the function
f
(
x
)
=
3
x
−
4
x
f(x)=\dfrac{3x-4}{x}
f
(
x
)
=
x
3
x
−
4
. Find the second derivative
f
′
′
(
x
)
,
at
x
=
1
f''(x),\text{ at }x=1
f
′′
(
x
)
,
at
x
=
1
.
Higher Order Derivatives
An object is thrown straight up in the air at
t
=
0
t = 0
t
=
0
seconds. Its height in metres at
t
t
t
seconds is given by
h
(
t
)
=
s
0
+
v
0
t
−
7
t
2
h(t) = s_0 + v_0 t - 7t^2
h
(
t
)
=
s
0
+
v
0
t
−
7
t
2
where
s
0
s_0
s
0
and
v
0
v_0
v
0
are constants. In the first second the object rises 7 metres. For how many seconds does the object rise before starting to fall back down?
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
Higher Order Derivatives: The Power Rule
If
f
(
x
)
=
x
10
f\left(x\right)=x^{10}
f
(
x
)
=
x
10
find
f
(
9
)
(
x
)
f^{\left(9\right)}\left(x\right)
f
(
9
)
(
x
)
.
(i.e. the 9th derivative)
Higher Order Derivatives
Find the third derivative of
f
(
x
)
=
x
2
+
16
x
f(x) = x^2+16\sqrt{x}
f
(
x
)
=
x
2
+
16
x
.
Higher Order Derivatives
If
f
(
x
)
=
sin
(
2
x
)
f\left(x\right)=\sin\left(2x\right)
f
(
x
)
=
sin
(
2
x
)
, find
f
(
21
)
(
π
2
)
f^{\left(21\right)}\left(\frac{\pi}{2}\right)
f
(
21
)
(
2
π
)
.
(i.e. find the 21st derivative at the point
π
2
\frac{\pi}{2}
2
π
)
Higher order derivatives
Consider the function
f
(
x
)
=
3
x
−
4
x
f(x)=\dfrac{3x-4}{x}
f
(
x
)
=
x
3
x
−
4
. Find the second derivative
f
′
′
(
x
)
,
at
x
=
1
f''(x),\text{ at }x=1
f
′′
(
x
)
,
at
x
=
1
.