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

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Answer 1 · 2015-10-22T17:19:35

$sec x$

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

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

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

$=sec x$

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