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High School Math : Finding Derivatives

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Example Question #1 : Specific Derivatives

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

Possible Answers:

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

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

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

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

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 #2 : Specific Derivatives

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

Possible Answers:

$-\frac{ 1}{27}$

$\frac{ 1}{27}$

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

$-27\ln 3$

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

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 #3 : Specific Derivatives

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

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 #2141 : High School Math

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

What is f'(4) ?

Possible Answers:

$\ln 36$

$6 \ln 6$

$36 \ln 6$

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

$\ln 6$

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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Example Question #1 : Understanding Derivatives Of Trigonometric Functions

Find the derivative of the following function:

$f(x) = 3 \;\textup{sin}(x)$

Possible Answers:

$f(x) = 3 \;\textup{cos}(x)$

$f(x) = 3 \;\textup{tan}(x)$

$f(x) = 3 \;\textup{csc}(x)$

$f(x) = -3 \;\textup{sin}(x)$

$f(x) = -3 \;\textup{cos}(x)$

Correct answer:

$f(x) = 3 \;\textup{cos}(x)$

Explanation:

The derivative of $\sin (x)$ is $\cos (x)$ . It is probably best to memorize this fact (the proof follows from the difference quotient definition of a derivative).

Our function

$f(x) = 3 \;\textup{sin}(x)$

the factor of 3 does not change when we differentiate, therefore the answer is

$f(x) = 3 \;\textup{cos}(x)$

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Example Question #1 : Understanding The Derivative Of Trigonometric Functions

$\frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=?$

Possible Answers:

$\cos(x)$

$\sec(x)$

$\csc(x)$

$-\sin(x)$

$\sin(x)$

Correct answer:

$\cos(x)$

Explanation:

The derivative of a sine function does not follow the power rule. It is one that should be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

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Example Question #3 : Understanding Derivatives Of Trigonometric Functions

What is the second derivative of $\sin(x)$ ?

Possible Answers:

$\csc(x)$

$\cos(x)$

$-\sin(x)$

$\sec(x)$

$-\cos(x)$

Correct answer:

$-\sin(x)$

Explanation:

The derivatives of trig functions must be memorized. The first derivative is:

$\frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)$ .

To find the second derivative, we take the derivative of our result.

$\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-\sin(x)$ .

Therefore, the second derivative will be $-\sin(x)$ .

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Example Question #5 : Understanding The Derivative Of Trigonometric Functions

Compute the derivative of the function $f(x) = \tan x^{2}$ .

Possible Answers:

$f' (x) = x^{2} \tan x^{2} + 2x \sec^{2} x^{2}$

$f' (x) = x^{2} \sec^{2} x^{2}$

$f' (x) = 2x \sec x^{2}$

$f' (x) = 2x \sec^{2} x^{2}$

$f' (x) = x^{2} \sec 2x$

Correct answer:

$f' (x) = 2x \sec^{2} x^{2}$

Explanation:

Use the Chain Rule.

Set $u = x^{2}$ and substitute.

$f(x) = \tan u$

$\frac{\mathrm{d} f }{\mathrm{d} u} = \frac{\mathrm{d} }{\mathrm{d} u} \tan u = \sec^{2} u$

$\frac{\mathrm{d} u }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} x^{2} = 2x^{2-1} =2x$

$\frac{\mathrm{d} f }{\mathrm{d} x} = \frac{\mathrm{d} f }{\mathrm{d} u} \cdot \frac{\mathrm{d} u }{\mathrm{d} x}= \sec^{2} u \cdot 2x = 2x \sec^{2} x^{2}$

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Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Find the derivative of the following function:

$f(x) = x^3 + \frac{1}{x}$

Possible Answers:

$f'(x) = 3x^2-\frac{1}{x}$

f'(x) = 3x^2+x

$f'(x) = 3x^2+\frac{1}{x^2}$

f'(x) = 3x^2

$f'(x) = 3x^2-\frac{1}{x^2}$

Correct answer:

$f'(x) = 3x^2-\frac{1}{x^2}$

Explanation:

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of

x^3

is simply

3x^2

We can rewrite $\frac{1}{x}$ as

$x^{-1}$

and using the power rule again, we get a derivative of

$-x^{-2}$ or $-\frac{1}{x^2}$

So the answer is

$f'(x) = 3x^2-\frac{1}{x^2}$

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Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

What is $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} ?$

Possible Answers:

f(x)*f'(x)+g(x)*g'(x)

f(x)g'(x)-f'(x)g(x)

f(x)g(x)+f'(x)g'(x)

$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

f(x)g'(x)+f'(x)g(x)

Correct answer:

f(x)g'(x)+f'(x)g(x)

Explanation:

The chain rule is "first times the derivative of the second plus second times derivative of the first".

In this case, that means $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} =f(x)g'(x)+f'(x)g(x)$ .

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High School Math : Finding Derivatives

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Specific Derivatives

Example Question #2 : Specific Derivatives

Example Question #3 : Specific Derivatives

Example Question #2141 : High School Math

Example Question #1 : Understanding Derivatives Of Trigonometric Functions

Example Question #1 : Understanding The Derivative Of Trigonometric Functions

Example Question #3 : Understanding Derivatives Of Trigonometric Functions

Example Question #5 : Understanding The Derivative Of Trigonometric Functions

Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

All High School Math Resources

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