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

1 Answer
Jan 29, 2016

Answer:

#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))))#