Question

How do you find the derivative `y=sinh^-1(tanx)`?

Answer 1

`sec x`

Explanation:

From rules of differentiation for inverse hyperbolic trig functions and normal trig functions, we get

`d/dxsinh^(-1)(tanx)=1/(sqrt(1+tan^2x))*sec^2x`

`=1/sqrt(sec^2x)*sec^2x`

`=sec x`