differentiate arsinh(cosx))

let's start by defining y = arsinh(cos(x)). taking sinh of both sides gives sinhy = cosx. (since sinh(arsinhz) = z). Now we can differentiate both sides wrt x. The RHS differentiates to -sinx. We can use the chain rule for the LHS: d/dx = dy/dx *d/dy.so d/dx(sinhy) = dy/dx d/dy(sinhy) = dy/dx coshy. so dy/dx = -sinx/coshy. Now coshy = sqrt(1+(sinhy)^2) = sqrt(1+(cosx)^2).So dy/dx = -sinx/sqrt(1+(cosx)^2).

AB
Answered by Amit B. Further Mathematics tutor

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