elvishwitchxyp

2021-12-31

Finding the derivative of $\mathrm{cos}\left(\mathrm{arcsin}x\right)$

### Answer & Explanation

limacarp4

With your notation, $x=\mathrm{sin}u$, we have
$\frac{d}{dx}\left[\mathrm{cos}\left({\mathrm{sin}}^{-1}x\right)\right]=-\mathrm{sin}\left({\mathrm{sin}}^{-1}x\right)\cdot \frac{d}{dx}\left[{\mathrm{sin}}^{-1}x\right]=-x\frac{d}{dx}\left[{\mathrm{sin}}^{-1}x\right]$
Since
$\frac{dx}{du}=\mathrm{cos}u$
we have
$\frac{du}{dx}=\frac{1}{\mathrm{cos}u}=\frac{1}{\sqrt{1-{\mathrm{sin}}^{2}u}}=\frac{1}{\sqrt{1-{x}^{2}}}$
Therefore,
$\frac{d}{dx}\left[\mathrm{cos}\left({\mathrm{sin}}^{-1}x\right)\right]=-\frac{x}{\sqrt{1-{x}^{2}}}$

kalfswors0m

Let $y=\mathrm{cos}\left(u\right)=\mathrm{cos}\left(\mathrm{arcsin}x\right)$. 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 textbooks

Vasquez

Alternative approach:
The range of , so the $\mathrm{cos}\left[\mathrm{arcsin}\left(x\right)\right]$ will be $\ge 0$
Further, if $\theta =\mathrm{arcsin}\left(x\right)$, then $\mathrm{cos}\left(\theta \right)$ [which by the above statement will be non-negative] will be $\sqrt{1-{x}^{2}}$
Therefore, the question immediately reduces to computing
$\frac{d}{dx}\sqrt{1-{x}^{2}}$

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