Derivative of $\cos^{-1}\sqrt{\frac{1+x}2}$ using substitution

Question

Find the derivative of $$\cos^{-1}\sqrt{\frac{1+x}2}.$$

I'm learning differentiation and calculus for the first time. I can easily find the derivative of the given expression by chain rule. But the book from which I'm learning calculus encourages finding derivatives of inverse trigonometric functions of algebraic functions with substitution rather than using chain rule. So, I want to find the derivative of this function with substitution. Here is my attempt to do that:

Let $x=\cos2\theta$, then $\theta=\frac{\cos^{-1}x}2$.
Now,
$\begin{align}\cos^{-1}\sqrt{\frac{1+x}2} &= \cos^{-1}\left(\frac1{\sqrt2}\sqrt{1+\cos2\theta}\right)\\ &= \cos^{-1}\left(\frac1{\sqrt2}\sqrt{1+\cos^2\theta-1}\right)\\ &= \cos^{-1}\left(\frac1{\sqrt2}\cos\theta\right) \end{align}$
I can't proceed further from here. Can we write this as $\frac1{\sqrt2}\theta$? And please do not give a solution using chain rule, as this would be of no help to me.

Answer 1

Setting $x = \cos 2\theta$ and remembering that $\cos 2\theta = \color{red}{2}\cos^2\theta - 1$ leads to:

$$\begin{align} \cos^{-1}\sqrt{\frac{1+x}2} &= \cos^{-1}\sqrt{\frac{1+\cos2\theta}2} \\ &= \cos^{-1}\sqrt{\frac{1+\color{red}{2}\cos^2\theta - 1}2} \\ &= \theta \end{align}$$

And so one can use:

$$\frac{d}{dx}\cos^{-1}\sqrt{\frac{1+x}2} = \frac{d\theta}{dx} = \frac{1}{dx/d\theta}$$

Answer 2

Second last step is wrong.

You used the right substitution. Simply write $1+\cos2\theta=2\cos^2\theta$. Afterwards, there's nothing much left in the question.

Answer 3

Let $x=\cos (2\theta)$ then

$$\cos^{-1}\sqrt{\frac{1+x}2}=\cos^{-1}(\cos \theta)=\theta$$

and

$$\frac{dx}{d\theta}=-2\sin (2\theta)$$

$$\frac{d}{dx}\left(\cos^{-1}\sqrt{\frac{1+x}2}\right)=-\frac12 \frac1{\sin (2\theta)}=-\frac12 \frac1{\sqrt{1-x^2}}$$

Answer 4

I think the natural way should be:

Let $\cos^{-1}\sqrt{\dfrac{1+x}2}=y$

$\implies0\le y\le\dfrac\pi2$ for real $1\ge x\ge-1$

$\implies x=\cdots=\cos2y$

and consequently $2y=\cos^{-1}x$ as $0\le \cos^{-1}x\le\pi$