# Derivative of a sinusoidal function

## Homework Statement

What is the derivative of ##f(x)=\frac {2x^2} {cos x}##?

## The Attempt at a Solution

##F(x)=\frac {2x^2} {cos x}##

So...

##f(x)=2x^2## and ##f'(x)=4x##

##g(x)=cosx## and ##g'(x)=-sinx##

If I plug these into the quotient rule I thought that I would get the correct answer.

##F'(x)=\frac {f'(x)g(x)-f(x)g'(x)} {(g(x))^2}##

##=\frac {4x cosx -2x^2 (-sinx)} {cos^2x}##

The answer i'm given is supposed to be ##=\frac {4x cosx +2x^2 sinx} {cos^2x}##.

What is happening to make ##-2x^2## positive and ##-sinx## positive?

berkeman
Mentor
The answers look the same to me. (-1)*(-1) is positive...

• DaveE and TachyonLord
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

What is the derivative of ##f(x)=\frac {2x^2} {cos x}##?

## The Attempt at a Solution

##F(x)=\frac {2x^2} {cos x}##

So...

##f(x)=2x^2## and ##f'(x)=4x##

##g(x)=cosx## and ##g'(x)=-sinx##

If I plug these into the quotient rule I thought that I would get the correct answer.

##F'(x)=\frac {f'(x)g(x)-f(x)g'(x)} {(g(x))^2}##

##=\frac {4x cosx -2x^2 (-sinx)} {cos^2x}##

The answer i'm given is supposed to be ##=\frac {4x cosx +2x^2 sinx} {cos^2x}##.

What is happening to make ##-2x^2## positive and ##-sinx## positive?

I assume you mean ##\cos x## rather than ##cos x,## and to get the first form, just put a "\" in front (so type "\cos x" instead of "cos x"). Same for sin, tan, etc---basically for all the standard functions.
Anyway, the quotient rule gives
$$\left( \frac{2x^2}{\cos x} \right)'= \frac{(2x^2)'}{\cos x} - \frac{(2 x^2) (\cos x)'}{\cos^2 x},$$
which evaluates to
$$\frac{4x\: \cos x -2x^2 \: (-\sin x)}{\cos^2 x} = \text{given answer}$$ Use the fact that ##-(- \sin x) = + \sin x##.

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