Understanding Derivatives of Exponents - High School Math

Card 0 of 4

Question

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

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 $2^{x+1}$

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Question

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

What is ?

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 ?

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 .

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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