Question

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

Answer 1

`(2x)/(1-x^2)cot(ln(1-x^2))csc(ln(1-x^2))`

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(g(h(x)))`. But the same rule still applies.

Let us define:
`f(x) = csc(x)`
`h(x) - ln(x)`
`g(x) = 1-x^2`

Let's start by looking at the inner most function: `h(x)=(1-x^2)`.
Differentiating this yields `d/dxh(x) = -2x`.

Now let's look at `g(h(x)) = ln(1-x^2)`

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

gives us `d/dxg(h(x)) =h'(x)1/(h(x))`.

Substituting `h(x)` in gives us: `d/dxg(h(x)) = -(2x)/(1-x^2)`

Finally lets look at `f(g(h(x)))`

Note that `d/dx csc(x) = -cot(x)csc(x)`

So differentiating `f(g(h(x))) = csc(g(h(x)))` 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:

`d/dxcsc(ln(1-x^2)) = (2x)/(1-x^2)cot(ln(1-x^2))csc(ln(1-x^2))`

And that is our final answer.