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How do you find the derivative of the function: #Sin(Arc Cosx)#?

Answer 1

Emma Aldridge

I would rewrite using trigonometry, then differentiate

#theta = arccosx# if and only if #0 <= theta <= pi# and #cos theta = x#

Therefore #sin theta = sqrt(1-cos^2theta) = sqrt(1-x^2)#

So,

#d/dx(sin(arccos(x))) = d/dx(sqrt(1-x^2))#

# = 1/(2sqrt(1-x^2)) d/dx(1-x^2)#

(using #d/dx(sqrtu) = 1/(2sqrtu) d/dx(u)#)

# = 1/(2sqrt(1-x^2)) (-2x)#

# = -x/sqrt(1-x^2)#

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Answer 2

Charles Arocha

To find the derivative of the function sin(arc cos x), you can use the chain rule. Let y = sin(arc cos x). Then differentiate y with respect to x using the chain rule:

dy/dx = (dy/du) * (du/dx)

where u = arc cos x.

Differentiating sin(u) with respect to u gives cos(u), and differentiating arc cos x with respect to x gives -1 / sqrt(1 - x^2).

Therefore,

dy/dx = cos(arc cos x) * (-1 / sqrt(1 - x^2))

Simplify cos(arc cos x) to just x:

dy/dx = x * (-1 / sqrt(1 - x^2))

So, the derivative of sin(arc cos x) with respect to x is:

dy/dx = -x / sqrt(1 - x^2)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.