What is the derivative of this function #y=sec^-1(4x)#?

1 Answer
Oct 6, 2016

#dy/dx=1/(absxsqrt(16x^2-1))#

Explanation:

You may already know that #d/dxsec^-1(x)=1/(absxsqrt(x^2-1))#. Then, according to the chain rule, we see that we will be following the rule:

#d/dxsec^-1(f(x))=1/(abs(f(x))sqrt((f(x))^2-1))*f'(x)#

So, for #sec^-1(4x)#, we see that:

#dy/dx=1/(abs(4x)sqrt((4x)^2-1))*d/dx4x#

#dy/dx=4/(abs(4x)sqrt(16x^2-1))#

Note that #abs(4x)=4absx#:

#dy/dx=1/(absxsqrt(16x^2-1))#