# How do you differentiate y=e^(ktansqrtx)?

Derivative = k/2 x^(-1/2) sec^2(sqrtx)e^(ktan(sqrtx)
If $f \left(x\right) = {e}^{g \left(x\right)}$ then $f ' \left(x\right) = g ' \left(x\right) {e}^{g \left(x\right)}$