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.
Best Answer
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, $$\frac{d}{dx}\left[\cos(\sin^{-1} x)\right] = - \frac{x}{\sqrt{1-x^2}}.$$