High School Math : Derivatives

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #1 : Finding First And Second Derivatives

Find the second derivative of f(x).

 

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

Possible Answers:

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

\displaystyle e^{2x}+cosx

\displaystyle e^{2x}+sinx

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

\displaystyle e^{2x}-cosx

Correct answer:

\displaystyle e^{2x}+cosx

Explanation:

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

 

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

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

 

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

 

 

Example Question #2 : Finding First And Second Derivatives

Find the derivative of the function

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We can use the Chain Rule:

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

\displaystyle \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

 

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

Evaluate the following limit:

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

Possible Answers:

\displaystyle -1

\displaystyle 0

\displaystyle 2

\displaystyle 1

\displaystyle \infty

Correct answer:

\displaystyle 0

Explanation:

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

 

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

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