Derivatives - High School Math

Card 0 of 3

Question

Find the second derivative of f(x).

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

Answer

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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Question

Find the derivative of the function

.

Answer

We can use the Chain Rule:

Let , so that .

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

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Question

Evaluate the following limit:

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

Answer

When approaches 0 both and 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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