Derivative Review

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Math › Derivative Review

Questions 1 - 4

1

Find the derivative of the function

.

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

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

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

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

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

Explanation

We can use the Chain Rule:

Let , so that .

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

2

Find the derivative of the function

.

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

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

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

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

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

Explanation

We can use the Chain Rule:

Let , so that .

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

3

Find the second derivative of f(x).

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

$e^{2x}+cosx$

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

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

$e^{2x}-cosx$

$e^{2x}+sinx$

Explanation

First we should find the first derivative of . Remember the derivative of $e^{2x}$ is $2e^{2x}$ and the derivative of is :

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

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

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

4

Find the second derivative of f(x).

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

$e^{2x}+cosx$

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

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

$e^{2x}-cosx$

$e^{2x}+sinx$

Explanation

First we should find the first derivative of . Remember the derivative of $e^{2x}$ is $2e^{2x}$ and the derivative of is :

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

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

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

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