Best solutions to - "Finding the derivative of cos⁡(arcsin⁡x)"

elvishwitchxyp

Answered question

2021-12-31

Finding the derivative of

\cos (\arcsin x)

Answer & Explanation

limacarp4

Beginner2022-01-01Added 39 answers

With your notation,

x = \sin u

, we have

\frac{d}{d x} [\cos (\sin^{- 1} x)] = - \sin (\sin^{- 1} x) \cdot \frac{d}{d x} [\sin^{- 1} x] = - x \frac{d}{d x} [\sin^{- 1} x]

Since

\frac{d x}{d u} = \cos u

we have

\frac{d u}{d x} = \frac{1}{\cos u} = \frac{1}{\sqrt{1 - \sin^{2} u}} = \frac{1}{\sqrt{1 - x^{2}}}

Therefore,

\frac{d}{d x} [\cos (\sin^{- 1} x)] = - \frac{x}{\sqrt{1 - x^{2}}}

kalfswors0m

Beginner2022-01-02Added 24 answers

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 textbooks

Vasquez

Expert2022-01-08Added 669 answers

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

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