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

1 Answer
Jim H
Jan 29, 2016

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

Explanation:

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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