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High School Math : Understanding Derivatives of Exponents

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Example Questions

Example Question #51 : Finding Derivatives

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

Possible Answers:

$x2^{x+1}+\ln(2)x^{2}2^{x}$

$2x2^{x}-x^22^{x-1}$

$2x\ln(2)2^{x}$

$2x2^{x}+x^{3}2^{x-1}$

Correct answer:

$x2^{x+1}+\ln(2)x^{2}2^{x}$

Explanation:

The derivative must be computed using the product rule. Because the derivative of x^2 brings a down as a coefficient, it can be combined with 2^x to give $2^{x+1}$

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Example Question #1 : Specific Derivatives

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

Possible Answers:

$-27\ln 3$

$-\frac{ 1}{27}$

$\frac{ \ln 3}{27}$

$\frac{ 1}{27}$

$-\frac{ \ln 3}{27}$

Correct answer:

$-\frac{ \ln 3}{27}$

Explanation:

The instantaneous rate of change of at $x = x _{0}$ is $f ' (x _{0})$ , so we will find f ' (x) and evaluate it at x = 3 .

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ 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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Example Question #2 : Understanding Derivatives Of Exponents

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

What is f'(x) ?

Possible Answers:

$f'(x) = \frac{5 ^{x}}{\ln 5}$

$f'(x) = 5 ^{x}$

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

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

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

Correct answer:

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

Explanation:

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

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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Example Question #1 : Specific Derivatives

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

What is f'(4) ?

Possible Answers:

$\frac{\ln 6}{6}$

$\ln 6$

$6 \ln 6$

$\ln 36$

$36 \ln 6$

Correct answer:

$36 \ln 6$

Explanation:

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

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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High School Math : Understanding Derivatives of Exponents

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #51 : Finding Derivatives

Example Question #1 : Specific Derivatives

Example Question #2 : Understanding Derivatives Of Exponents

Example Question #1 : Specific Derivatives

All High School Math Resources

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