What is the derivative of sin x / cos ^2 x - 1 ?

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What is the derivative of sin x / cos ^2 x - 1 ?

I come to the answer $\frac{- cos x {sin}^{2} x}{{({cos}^{2} x - 1)}^{2}}$ $(-cos x sin^2 x)/((cos^2x -1)^2)$

But it should be :
$\frac{- cos x}{({cos}^{2} x - 1)}$ $(-cos x )/((cos^2x -1))$

I know ${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x = 1 - cos^2x$

But its not the same as ${cos}^{2} x - 1$ $cos^2x -1$

So I do something wrong.

My first step was :

$\frac{(- 2 cos x sin x sin x) - (cos x ({cos}^{2} x - 1))}{{({cos}^{2} x - 1)}^{2}}$ $((-2 cos x sinx sinx ) - (cos x (cos^2x -1)))/ (cos^2x -1)^2$

Simplify
$\frac{(- 2 cos x {sin}^{2} x) - cos x ({cos}^{2} x - 1)}{{({cos}^{2} x - 1)}^{2}}$ $((-2cosx sin^2x) - cos x (cos^2x-1))/((cos^2x -1)^2)$

Gives:
$\frac{- cos x (({cos}^{2} x - 1) + 2 {sin}^{2} x)}{{({cos}^{2} x - 1)}^{2}}$ $(-cosx((cos^2x-1)+2 sin^2x))/((cos^2x -1)^2)$

Gives:
$\frac{- cos x (1 + {sin}^{2} x - 1)}{{({cos}^{2} x - 1)}^{2}}$ $(-cosx(1+sin^2x-1))/((cos^2x -1)^2)$

Gives:
$\frac{- cos x {sin}^{2} x}{{({cos}^{2} x - 1)}^{2}}$ $(-cosxsin^2x)/((cos^2x -1)^2)$

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Answer 1 · 2018-04-17T20:28:18

You're almost there!

What you have to realize is that ${cos}^{2} x - 1 = - (1 - {cos}^{2} x)$ $cos^2x-1=-(1-cos^2x)$ . Try distributing the minus sign and see what you get.

Once you understand that, it's just one step further to say that ${cos}^{2} x - 1 = - {sin}^{2} x$ $cos^2x-1=-sin^2x$ . You can also get this equation straight away by moving terms around in ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ (move the $1$ $1$ and ${sin}^{2} x$ $sin^2x$ to their opposite sides).

Another error you have is that you reversed the order of the terms in the numerator when you did the quotient rule, remember, $\frac{d}{d x} (\frac{u}{v}) = \frac{\frac{d u}{d x} \cdot v - u \cdot \frac{d v}{d x}}{v^{2}}$ $d/dx(u/v)=((du)/dx*v-u*(dv)/dx)/v^2$ .

This makes what you got off by a factor of $- 1$ $-1$ , so you should've gotten just $\frac{cos x {sin}^{2} x}{{({cos}^{2} x - 1)}^{2}}$ $(cosxsin^2x)/(cos^2x-1)^2$ as your answer (I took away the minus sign).

From here, use the identity ${cos}^{2} x - 1 = - {sin}^{2} x$ $cos^2x-1=-sin^2x$ :

$\frac{cos x {sin}^{2} x}{{({cos}^{2} x - 1)}^{2}} = \frac{cos x {sin}^{2} x}{({cos}^{2} x - 1) ({cos}^{2} x - 1)}$ $(cosxsin^2x)/(cos^2x-1)^2=(cosxsin^2x)/((cos^2x-1)(cos^2x-1))$

$= \frac{cos x {sin}^{2} x}{{cos}^{2} x - 1} (- {sin}^{2} x))$ $=(cosxsin^2x)/(cos^2x-1)(-sin^2x))$

$= \frac{- cos x}{{cos}^{2} x - 1}$ $=(-cosx)/(cos^2x-1)$

As you wished to show.

But, honestly, I think the best way to express this is by rewriting the remaining ${cos}^{2} x - 1$ $cos^2x-1$ :

$= \frac{- cos x}{- {sin}^{2} x}$ $=(-cosx)/(-sin^2x)$

$= \frac{cos x}{sin x} (\frac{1}{sin x})$ $=cosx/sinx(1/sinx)$

$= cot x csc x$ $=cotxcscx$

If you know your trigonometric derivatives, you might recognize that this is off the derivative of $csc x$ $cscx$ by a factor of $- 1$ $-1$ . This isn't surprising, though, when we consider the original function:

$\frac{sin x}{{cos}^{2} x - 1} = \frac{sin x}{- {sin}^{2} x} = \frac{- 1}{sin x} = - csc x$ $sinx/(cos^2x-1)=sinx/(-sin^2x)=(-1)/sinx=-cscx$

So the derivative is exactly as we'd expect.

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What is the derivative of sin x / cos ^2 x - 1 ?

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