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How do you find the derivative of the function: #sin(arccosx)#?

Answer 1

#d/dx(sin(arccosx)) = cos(arccosx)*d/dx(arccosx) = x * (-1)/sqrt(1-x^2)# # = (-x)/sqrt(1-x^2)#

It is possible to use the chain rule and the derivatives of #sin(u)# and #arccos(x)#, then use trigonometry to simplify. (As above.) But I prefer to do the trigonometry first.

#sin(arccosx)# is the sine of a number in #[0,pi]# whose cosine is #x#. (Definition of #arccosx#)

The sine of such an angle is #sqrt(1-x^2)#.

That is,

#sin(arccosx) = sqrt(1-x^2)#

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

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Answer 2

Cameron Alexander

To find the derivative of the function ( \sin(\arccos(x)) ), we use the chain rule. The derivative is given by ( -\frac{1}{\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.