Question

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

Answer 1

`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))`