Derivative of a sinusoidal function

[Specter]

Homework Statement

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

Homework Equations

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]

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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 $-(- \sin x) = + \sin x$.