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High School Math : Derivatives

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Example Questions

Example Question #1 : Derivatives

Find the second derivative of f(x).

$f(x)=\frac{1}{4} e^{2x}-cosx$

Possible Answers:

$e^{2x}+sinx$

$\frac{1}{2}e^{2x}+sinx$

$e^{2x}-cosx$

$\frac{1}{2}e^{2x}+cosx$

$e^{2x}+cosx$

Correct answer:

$e^{2x}+cosx$

Explanation:

First we should find the first derivative of f(x) . Remember the derivative of $e^{2x}$ is $2e^{2x}$ and the derivative of cosx is -sinx :

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

The second derivative is just the derivative of the first derivative:

$f''(x)=e^{2x}+cosx$

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Example Question #1 : Finding First And Second Derivatives

Find the derivative of the function

f(x)=tan^4x-tan^3x+tan^2x .

Possible Answers:

$(4tan^3x-3tan^2x+2tanx)\cdot sec^2x$

$\frac{4tan^3x+3tan^2x+2tanx}{sec^2x}$

$(4tan^3x-3tan^2x+2tanx)\cdot secx$

$\frac{4tan^3x-3tan^2x+2tanx}{csc^2x}$

$(4tan^3x-3tan^2x+2tanx)\cdot csc^2x$

Correct answer:

$(4tan^3x-3tan^2x+2tanx)\cdot sec^2x$

Explanation:

We can use the Chain Rule:

Let u=tanx , so that f(x)=u^4-u^3+u^2 .

$\frac{df}{dx}=\frac{df}{du}.\frac{du}{dx}=(4u^3-3u^2+2u)\cdot sec^2x=(4tan^3x-3tan^2x+2tanx)\cdot sec^2x$

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Example Question #1 : Understanding L'hospital's Rule

Evaluate the following limit:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})$

Possible Answers:

$\infty$

Correct answer:

Explanation:

When approaches 0 both sinx and (1-cos2x) will approach . Therefore, L’Hopital’s Rule can be applied here. Take the derivatives of the numerator and denominator and try the limit again:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})= \lim_{x\rightarrow 0} (\frac{2sin2x}{cosx})=\frac{0}{1}=0$

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High School Math : Derivatives

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Derivatives

Example Question #1 : Finding First And Second Derivatives

Example Question #1 : Understanding L'hospital's Rule

All High School Math Resources

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