Understanding Derivatives of Exponents - Math

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Question

Find the derivative for $y = $x^{2}$$2^x$$

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Answer

The derivative must be computed using the product rule. Because the derivative of brings a down as a coefficient, it can be combined with to give

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Question

$f(x) = 5 ^{x + 1}$

What is ?

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Answer

$f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }$

Therefore,

$f'(x) = 5 \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any real , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}$ , and

$f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}$

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Question

$f(x) = 6 ^{x - 2}$

What is ?

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Answer

$f(x) = 6 ^{x -2} = 6 ^{-2} \cdot 6 ^{x } = $\frac{1}{36}$ \cdot 6 ^{x }$

Therefore,

$f'(x) = $\frac{1}{36}$ \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right ) = \ln 6 \cdot 6 ^{x}$ , and

$f'(x) = $\frac{1}{36}$ \ln 6 \cdot 6 ^{x} = \ln 6 \cdot $\frac{1}{36}$ \cdot 6 ^{x}$

$f'(x) = \ln 6 \cdot 6 ^{x-2}$

$f'(4) = \ln 6 \cdot 6 ^{4-2} = \ln 6 \cdot 6 ^{2} = 36 \ln 6$

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Question

Give the instantaneous rate of change of the function $f (x) = \left ( $\frac{1}{3}$ \right ) ^{x}$ at .

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Answer

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}$}{\mathrm{d} x}\left[ \left( $\frac{1}{3}$ \right) ^{x }\right] = \ln $\frac{1}{3}$ \cdot \left( $\frac{1}{3}$ \right) ^{x }= -$\frac{ \ln $3}{3^{x}$$}$

$f'(3) =-$\frac{ \ln $3}{3^{3}$$} = -$\frac{ \ln 3}{27}$$

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Question

Find the derivative for $y = $x^{2}$$2^x$$

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Answer

The derivative must be computed using the product rule. Because the derivative of brings a down as a coefficient, it can be combined with to give

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Question

$f(x) = 5 ^{x + 1}$

What is ?

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Answer

$f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }$

Therefore,

$f'(x) = 5 \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any real , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}$ , and

$f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}$

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Question

$f(x) = 6 ^{x - 2}$

What is ?

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Answer

$f(x) = 6 ^{x -2} = 6 ^{-2} \cdot 6 ^{x } = $\frac{1}{36}$ \cdot 6 ^{x }$

Therefore,

$f'(x) = $\frac{1}{36}$ \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right ) = \ln 6 \cdot 6 ^{x}$ , and

$f'(x) = $\frac{1}{36}$ \ln 6 \cdot 6 ^{x} = \ln 6 \cdot $\frac{1}{36}$ \cdot 6 ^{x}$

$f'(x) = \ln 6 \cdot 6 ^{x-2}$

$f'(4) = \ln 6 \cdot 6 ^{4-2} = \ln 6 \cdot 6 ^{2} = 36 \ln 6$

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Question

Give the instantaneous rate of change of the function $f (x) = \left ( $\frac{1}{3}$ \right ) ^{x}$ at .

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Answer

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}$}{\mathrm{d} x}\left[ \left( $\frac{1}{3}$ \right) ^{x }\right] = \ln $\frac{1}{3}$ \cdot \left( $\frac{1}{3}$ \right) ^{x }= -$\frac{ \ln $3}{3^{x}$$}$

$f'(3) =-$\frac{ \ln $3}{3^{3}$$} = -$\frac{ \ln 3}{27}$$

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Question

Find the derivative for $y = $x^{2}$$2^x$$

Tap to reveal answer

Answer

The derivative must be computed using the product rule. Because the derivative of brings a down as a coefficient, it can be combined with to give

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Question

$f(x) = 5 ^{x + 1}$

What is ?

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Answer

$f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }$

Therefore,

$f'(x) = 5 \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any real , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}$ , and

$f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}$

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Question

$f(x) = 6 ^{x - 2}$

What is ?

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Answer

$f(x) = 6 ^{x -2} = 6 ^{-2} \cdot 6 ^{x } = $\frac{1}{36}$ \cdot 6 ^{x }$

Therefore,

$f'(x) = $\frac{1}{36}$ \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right ) = \ln 6 \cdot 6 ^{x}$ , and

$f'(x) = $\frac{1}{36}$ \ln 6 \cdot 6 ^{x} = \ln 6 \cdot $\frac{1}{36}$ \cdot 6 ^{x}$

$f'(x) = \ln 6 \cdot 6 ^{x-2}$

$f'(4) = \ln 6 \cdot 6 ^{4-2} = \ln 6 \cdot 6 ^{2} = 36 \ln 6$

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Question

Give the instantaneous rate of change of the function $f (x) = \left ( $\frac{1}{3}$ \right ) ^{x}$ at .

Tap to reveal answer

Answer

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}$}{\mathrm{d} x}\left[ \left( $\frac{1}{3}$ \right) ^{x }\right] = \ln $\frac{1}{3}$ \cdot \left( $\frac{1}{3}$ \right) ^{x }= -$\frac{ \ln $3}{3^{x}$$}$

$f'(3) =-$\frac{ \ln $3}{3^{3}$$} = -$\frac{ \ln 3}{27}$$

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Question

Find the derivative for $y = $x^{2}$$2^x$$

Tap to reveal answer

Answer

The derivative must be computed using the product rule. Because the derivative of brings a down as a coefficient, it can be combined with to give

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Question

$f(x) = 5 ^{x + 1}$

What is ?

Tap to reveal answer

Answer

$f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }$

Therefore,

$f'(x) = 5 \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any real , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}$ , and

$f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}$

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Question

$f(x) = 6 ^{x - 2}$

What is ?

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Answer

$f(x) = 6 ^{x -2} = 6 ^{-2} \cdot 6 ^{x } = $\frac{1}{36}$ \cdot 6 ^{x }$

Therefore,

$f'(x) = $\frac{1}{36}$ \cdot $\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right )$

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so $$\frac{\mathrm{d}$ }{\mathrm{d} x} \left ( 6 ^{x } \right ) = \ln 6 \cdot 6 ^{x}$ , and

$f'(x) = $\frac{1}{36}$ \ln 6 \cdot 6 ^{x} = \ln 6 \cdot $\frac{1}{36}$ \cdot 6 ^{x}$

$f'(x) = \ln 6 \cdot 6 ^{x-2}$

$f'(4) = \ln 6 \cdot 6 ^{4-2} = \ln 6 \cdot 6 ^{2} = 36 \ln 6$

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Question

Give the instantaneous rate of change of the function $f (x) = \left ( $\frac{1}{3}$ \right ) ^{x}$ at .

Tap to reveal answer

Answer

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find and evaluate it at .

$$\frac{\mathrm{d}$}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any positive , so

$f'(x) =\frac{\mathrm{d}$}{\mathrm{d} x}\left[ \left( $\frac{1}{3}$ \right) ^{x }\right] = \ln $\frac{1}{3}$ \cdot \left( $\frac{1}{3}$ \right) ^{x }= -$\frac{ \ln $3}{3^{x}$$}$

$f'(3) =-$\frac{ \ln $3}{3^{3}$$} = -$\frac{ \ln 3}{27}$$

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