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

1 Answer
Feb 5, 2016

Answer:

#(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.