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

## I come to the answer $\frac{- \cos x {\sin}^{2} x}{{\left({\cos}^{2} x - 1\right)}^{2}}$ But it should be : $\frac{- \cos x}{\left({\cos}^{2} x - 1\right)}$ I know ${\sin}^{2} x = 1 - {\cos}^{2} x$ But its not the same as ${\cos}^{2} x - 1$ So I do something wrong. My first step was : $\frac{\left(- 2 \cos x \sin x \sin x\right) - \left(\cos x \left({\cos}^{2} x - 1\right)\right)}{{\cos}^{2} x - 1} ^ 2$ Simplify $\frac{\left(- 2 \cos x {\sin}^{2} x\right) - \cos x \left({\cos}^{2} x - 1\right)}{{\left({\cos}^{2} x - 1\right)}^{2}}$ Gives: $\frac{- \cos x \left(\left({\cos}^{2} x - 1\right) + 2 {\sin}^{2} x\right)}{{\left({\cos}^{2} x - 1\right)}^{2}}$ Gives: $\frac{- \cos x \left(1 + {\sin}^{2} x - 1\right)}{{\left({\cos}^{2} x - 1\right)}^{2}}$ Gives: $\frac{- \cos x {\sin}^{2} x}{{\left({\cos}^{2} x - 1\right)}^{2}}$

Apr 17, 2018

You're almost there!

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

Once you understand that, it's just one step further to say that ${\cos}^{2} x - 1 = - {\sin}^{2} x$. You can also get this equation straight away by moving terms around in ${\sin}^{2} x + {\cos}^{2} x = 1$ (move the $1$ and ${\sin}^{2} x$ 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, $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{\frac{\mathrm{du}}{\mathrm{dx}} \cdot v - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$.

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

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

$\frac{\cos x {\sin}^{2} x}{{\cos}^{2} x - 1} ^ 2 = \frac{\cos x {\sin}^{2} x}{\left({\cos}^{2} x - 1\right) \left({\cos}^{2} x - 1\right)}$

=(cosxsin^2x)/(cos^2x-1)(-sin^2x))

$= \frac{- \cos x}{{\cos}^{2} x - 1}$

As you wished to show.

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

$= \frac{- \cos x}{- {\sin}^{2} x}$

$= \cos \frac{x}{\sin} x \left(\frac{1}{\sin} x\right)$

$= \cot x \csc x$

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

$\sin \frac{x}{{\cos}^{2} x - 1} = \sin \frac{x}{- {\sin}^{2} x} = \frac{- 1}{\sin} x = - \csc x$

So the derivative is exactly as we'd expect.