Finding the derivative of cos⁡(arcsin⁡x)

elvishwitchxyp

elvishwitchxyp

Answered

2021-12-31

Finding the derivative of cos(arcsinx)

Answer & Explanation

limacarp4

limacarp4

Expert

2022-01-01Added 39 answers

With your notation, x=sinu, we have
ddx[cos(sin1x)]=sin(sin1x)ddx[sin1x]=xddx[sin1x]
Since
dxdu=cosu
we have
dudx=1cosu=11sin2u=11x2
Therefore,
ddx[cos(sin1x)]=x1x2
kalfswors0m

kalfswors0m

Expert

2022-01-02Added 24 answers

Let y=cos(u)=cos(arcsinx). You have correctly worked out that du/dx is
11x2
but you need to multiply this result by dy/du. The textbooks
Vasquez

Vasquez

Expert

2022-01-08Added 457 answers

Alternative approach:
The range of arcsin is [π2,π2], so the cos[arcsin(x)] will be 0
Further, if θ=arcsin(x), then cos(θ) [which by the above statement will be non-negative] will be 1x2
Therefore, the question immediately reduces to computing
ddx1x2

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