# How do you differentiate arcsin(csc(4/x)) ) using the chain rule?

##### 1 Answer
Feb 3, 2016

f'(x)=(4csc(4/x)cot(4/x))/(x^2sqrt(1-csc^2(4/x))

#### Explanation:

To differentiate an $\arcsin$ function, use the chain rule:

$\frac{d}{\mathrm{dx}} \left[\arcsin \left(u\right)\right] = \frac{1}{\sqrt{1 - {u}^{2}}} \cdot u '$

Here, $u = \csc \left(\frac{4}{x}\right)$, so we see that

$f ' \left(x\right) = \frac{1}{\sqrt{1 - {\csc}^{2} \left(\frac{4}{x}\right)}} \cdot \frac{d}{\mathrm{dx}} \left[\csc \left(\frac{4}{x}\right)\right]$

Now, to differentiate the $\csc$ function, use the chain rule again:

$\frac{d}{\mathrm{dx}} \left[\csc \left(u\right)\right] = - \csc \left(u\right) \cot \left(u\right) \cdot u '$

Since this time $u = \frac{4}{x}$,

$f ' \left(x\right) = \frac{1}{\sqrt{1 - {\csc}^{2} \left(\frac{4}{x}\right)}} \cdot - \csc \left(\frac{4}{x}\right) \cot \left(\frac{4}{x}\right) \cdot \frac{d}{\mathrm{dx}} \left[\frac{4}{x}\right]$

Use the product rule to see that $\frac{d}{\mathrm{dx}} \left[\frac{4}{x}\right] = \frac{d}{\mathrm{dx}} \left[4 {x}^{-} 1\right] = - 4 {x}^{-} 2 = - \frac{4}{x} ^ 2$, giving a final simplified derivative of

f'(x)=(4csc(4/x)cot(4/x))/(x^2sqrt(1-csc^2(4/x))