Question

How do you differentiate `f(x)=arccos(tan(1/(1-x^2)) )` using the chain rule?

Answer 1

`f'(x)=-(2xsec^2(1/(1-x^2)))/((1-x^2)^2sqrt(1-tan^2(1/(1-x^2))))`

Explanation:

To differentiate `arccos` functions, we can use the chain rule to state that

`d/dxarccos(u)=-1/sqrt(1-u^2)*u'`

Here, `u=tan(1/(1-x^2))`, so

`f'(x)=-1/sqrt(1-tan^2(1/(1-x^2)))*d/dxtan(1/(1-x^2))`

Now, to differentiate the `tan` function, we use the chain rule again:

`d/dxtan(u)=sec^2(u)*u'`

This time, `u=1/(1-x^2)=(1-x^2)^-1`, giving

`f'(x)=-1/sqrt(1-tan^2(1/(1-x^2)))*sec^2(1/(1-x^2))d/dx(1-x^2)^-1`

Before we find the next derivative, we can simplify a little.

`f'(x)=-sec^2(1/(1-x^2))/sqrt(1-tan^2(1/(1-x^2)))*d/dx(1-x^2)^-1`

To find this final derivative, use the chain rule once more:

`d/dxu^-1=-u^-2*u'`

This time around, `u=1-x^2`. Thus,

`f'(x)=-sec^2(1/(1-x^2))/sqrt(1-tan^2(1/(1-x^2)))*-(1-x^2)^-2d/dx(1-x^2)`

Since `d/dx(1-x^2)=-2x`, this whole mess simplifies to be

`f'(x)=-(2xsec^2(1/(1-x^2)))/((1-x^2)^2sqrt(1-tan^2(1/(1-x^2))))`