How to find $ \frac{d (\tanh(kx))}{d x}=?$

Question

I am tried to resolve the problem $$ \frac{d (\tanh(kx))}{d x}=?$$ where $k$ is positive value.

I found one solution that is $$ \frac{d (\tanh(kx))}{d x}=\frac{k}{2\cosh^2(kx)}$$

Is it right? If is not true, could you give me the true solution. Thanks

Answer 1

No, it is not right.$$\frac{d(\tanh (kx))}{dx}=\frac{1}{\cosh^2(kx)}\cdot \frac{d(kx)}{dx}=\frac{k}{\cosh^2(kx)}.$$

Answer 2

Just the classical chain rule. $$y=\tanh\big(u(x)\big)$$ $$y'=\text{sech}^2\big(u(x)\big)\frac{du(x)}{dx}$$ I do not know where the $2$ comes from.