# How do you differentiate f(x)=csc(ln(1-x^2))  using the chain rule?

##### 1 Answer
Feb 5, 2016

$\frac{2 x}{1 - {x}^{2}} \cot \left(\ln \left(1 - {x}^{2}\right)\right) \csc \left(\ln \left(1 - {x}^{2}\right)\right)$

#### Explanation:

Essentially the chain rule rule states that d/dx(f(g(x)) =g'(x)f'(g(x)). So we have a function "inside" another function. We start by identifying the "inside" function, differentiate it and then multiply it by the derivative of the "outside" function.

This example is obviously a bit more complicated because what we have here is a function inside a of function inside another function, or more formally: $f \left(g \left(h \left(x\right)\right)\right)$. But the same rule still applies.

Let us define:
$f \left(x\right) = \csc \left(x\right)$
$h \left(x\right) - \ln \left(x\right)$
$g \left(x\right) = 1 - {x}^{2}$

Let's start by looking at the inner most function: $h \left(x\right) = \left(1 - {x}^{2}\right)$.
Differentiating this yields $\frac{d}{\mathrm{dx}} h \left(x\right) = - 2 x$.

Now let's look at $g \left(h \left(x\right)\right) = \ln \left(1 - {x}^{2}\right)$

Differentiating the logarithm and applying the chain rule to g(h(x)) = ln(h(x))

gives us $\frac{d}{\mathrm{dx}} g \left(h \left(x\right)\right) = h ' \left(x\right) \frac{1}{h \left(x\right)}$.

Substituting $h \left(x\right)$ in gives us: $\frac{d}{\mathrm{dx}} g \left(h \left(x\right)\right) = - \frac{2 x}{1 - {x}^{2}}$

Finally lets look at $f \left(g \left(h \left(x\right)\right)\right)$

Note that $\frac{d}{\mathrm{dx}} \csc \left(x\right) = - \cot \left(x\right) \csc \left(x\right)$

So differentiating $f \left(g \left(h \left(x\right)\right)\right) = \csc \left(g \left(h \left(x\right)\right)\right)$ gives us

d/dxcsc(g(h(x))) = -d/dx{g(h(x))} cot(g(h(x))csc(g(h(x))

d/dx(g(h(x)) has already been obtained above.
Now substituting in the functions obtained previously we finally end up with:

$\frac{d}{\mathrm{dx}} \csc \left(\ln \left(1 - {x}^{2}\right)\right) = \frac{2 x}{1 - {x}^{2}} \cot \left(\ln \left(1 - {x}^{2}\right)\right) \csc \left(\ln \left(1 - {x}^{2}\right)\right)$

And that is our final answer.