Let $g(x)=\arccos(1/\cosh(5x))$. Find $g′(\ln(3))$

Question

Task

Let $g(x)$ be given by $g(x)=\arccos\left(\frac{1}{\cosh(5x)}\right)$.

Find $g′(ln(3))$.

I tried to start off with the formula

$\left(g^{-1}(x)\right)'=\frac{1}{g'(g^{-1}(x))}$

$\left(\arccos\left(\frac{1}{\cosh(5x)}\right)\right)'=\frac{1}{\sin\left(\arccos\left(\frac{1}{\cosh(5x)}\right)\right)}$

I am not allowed to post pictures yet, but it's just the triangle to help me find sin and arccos https://i.stack.imgur.com/HePn7.png

which tells us that $\sin\left(\arccos\left(\frac{1}{\cosh(5x)}\right)\right)=\frac{\sqrt{\cosh^2(5x)-1}}{\cosh(5x)}$

Which would give us $\left(g^{-1}(x)\right)'=\frac{\cosh(5x)}{\sqrt{\cosh^2(5x)-1}}$

But this is obviously not the right derivative of $g^{-1}(x)$

What did I do wrong and how do I solve the task?

Answer 1

We have, $$g(x) = \cos^{-1}\Bigg(\frac{1}{\cosh(5x)}\Bigg) = \cos^{-1}(\text{sech}(5x)).$$ Now, taking derivative with respect to $x$ we get,

$$g'(x) = \frac{5}{\sqrt{1-\text{sech}^2(5x)}}\tanh(5x)\text{sech}(5x)=5\text{sech}(5x).$$

Therefore, $$g'(\ln(3))=5\text{sech}(5\ln(3)) = \frac{10}{e^{5\ln(3)}+e^{-5\ln(3)}}=\frac{10}{3^5+3^{-5}}.$$

PS: Here I have used the following identities:

(i) $\frac{d\cos^{-1}(x)}{dx}=-\frac{1}{\sqrt{1-x^2}}$

(ii) $\frac{d\text{ sech}(x)}{dx}=-\tanh(x)\text{ sech}(x)$

(iii) $\tanh^2(x)+\text{sech}^2(x)=1$