elvishwitchxyp

Answered

2021-12-31

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

Answer & Explanation

limacarp4

Expert

2022-01-01Added 39 answers

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,

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

Since

we have

Therefore,

kalfswors0m

Expert

2022-01-02Added 24 answers

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

but you need to multiply this result by dy/du. The textbooks

Vasquez

Expert

2022-01-08Added 457 answers

Alternative approach:

The range of

Further, if

Therefore, the question immediately reduces to computing

Most Popular Questions