# How do you differentiate y=csc^-1x-4cot^-1x?

Feb 1, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{| x | \sqrt{{x}^{2} - 1}} + \frac{4}{1 + {x}^{2}}$

#### Explanation:

Step I

Let $\theta = {\csc}^{-} 1 x$
$\therefore x = \csc \theta$

I shall assume that there is no need to show the step by step differentiation of $\csc$, if I may (it just requires the chain rule, with $x = {\left(\sin \theta\right)}^{-} 1$, and skip straight to:

$\frac{\mathrm{dx}}{d \theta} = - \csc \theta \cot \theta$

We see that $\csc \theta$ is of course just $- x$. Then use the Pythagorean Identity to get $\cot \theta$ in terms of $x$.

$1 + {\cot}^{2} x = {\csc}^{2} x$
$\therefore \cot \theta = \sqrt{{x}^{2} - 1}$ (assume the positive square root here).

$\therefore \frac{\mathrm{dx}}{d \theta} = - x \sqrt{{x}^{2} - 1}$
:. (d theta)/dx = -1/(xsqrt(x^2-1)

Step II

Let $\alpha = {\cot}^{-} 1 x$
$\therefore x = \cot \alpha$

Again, if I may, I shall skip right to:

$\frac{\mathrm{dx}}{d \alpha} = - {\csc}^{2} \alpha$

Again, using the same Pythagorean Identity, we have:

$1 + {x}^{2} = {\csc}^{2} \alpha$
$\therefore - \left(1 + {x}^{2}\right) = \frac{\mathrm{dx}}{d \alpha}$
$\therefore \frac{d \alpha}{\mathrm{dx}} = - \frac{1}{1 + {x}^{2}}$

Step III

Now combine the lot:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{| x | \sqrt{{x}^{2} - 1}} + \frac{4}{1 + {x}^{2}}$, $x \in \left(- \infty , - 1\right) \cup \left(1 , \infty\right)$

Note: remember to re-introduce the constant $4$ which I left out whilst differentiating $4 {\cot}^{-} 1 x$

Second note: the absolute value bars rather mysteriously appeared on the denominator of that derivative. I believe this is because actually sketching the graph of $y = {\csc}^{-} 1 x$, the function is strictly decreasing, so by placing absolute value bars around the stray $x$, ensures that the derivative function stays negative for all $x$.

Third note: the domain comes in because the $\csc$ function is defined across this domain. The $\cot$ function is defined for all $x$, so the combined function takes the domain of the former.