Finding the derivative of $ \cos(\arcsin x)$ I study maths as a hobby. I am trying to find the derivative of $ \cos(\arcsin x)$
This is how I have been proceeding:
Let u = $\arcsin x$
Then $\sin u = x$
Differentiating:
\begin{align}
\cos u \frac{du}{dx} &= 1 \implies \frac{du}{dx} = \frac{1}{\cos u} \\[4pt]
\cos^2 u + \sin^2 u &= 1 \\[4pt]
\cos u &= \sqrt {1 - \sin^2 u} = \sqrt {1 - x^2}
\end{align}
But that is as far a I get.
The text book says the answer is $\frac{x}{\sqrt {1 - x^2}}$  but I cannot see how this is arrived at.
 A: With your notation, $x = \sin u$, we have $$\frac{d}{dx}\left[\cos (\sin^{-1} x)\right] = -\sin (\sin^{-1} x) \cdot \frac{d}{dx} \left[ \sin^{-1} x \right] = -x \frac{d}{dx} \left[ \sin^{-1} x \right].$$  Since $$\frac{dx}{du} = \cos u,$$ we have $$\frac{du}{dx} = \frac{1}{\cos u} = \frac{1}{\sqrt{1 - \sin^2 u}} = \frac{1}{\sqrt{1-x^2}}.$$  Therefore, $$\frac{d}{dx}\left[\cos(\sin^{-1} x)\right] = - \frac{x}{\sqrt{1-x^2}}.$$
A: Let $y=\cos(u)=\cos(\arcsin x)$. You have correctly worked out that $du/dx$ is
$$
\frac{1}{\sqrt{1-x^2}}
$$
but you need to multiply this result by $dy/du$. The textbook's answer actually contains a sign error.

This result can be obtained more directly if you write the chain rule in Lagrange notation. If $y=f(g(x))$, then
$$
\frac{dy}{dx}=f'(g(x)) \cdot g'(x) \, .
$$
Here, $f=\cos$ and $g=\arcsin$. Hence, $dy/dx$ equals
$$
-\sin(\arcsin(x)) \cdot\frac{1}{\sqrt{1-x^2}} \, ,
$$
which can be simplified further.
A: Alternative approach:
The range of ArcSin is $[-\pi/2, \pi/2]$, so the $\cos[$ArcSin$(x)]$ will be $\geq 0.$
Further, if $\theta = $ArcSin$(x)$, then $\cos(\theta)$ [which by the above statement will be non-negative] will be $\sqrt{1 - x^2}.$
Therefore, the question immediately reduces to computing
$$\frac{d}{dx} \sqrt{1 - x^2}.$$
