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How do you find the derivative of #Cos[arcsin(x)] #?

Answer 1

Emma Aldridge

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

Note that:

#cos^2 theta + sin^2 theta = 1#

So for any #theta#, we have:

#cos(theta) = +-sqrt(1-sin(theta))#

Note that for any real #x in [-1, 1]#, by definition:

#arcsin(x) in [-pi/2, pi/2]#

Further note that if #theta in [-pi/2, pi/2]# then:

#cos theta >= 0#

Hence:

#cos(arcsin(x)) = sqrt(1-x^2)#

for any #x in [-1, 1]#

Outside this interval #cos(arcsin(x))# takes non-real complex values.

So if we are dealing with this purely as a real valued function, then this is the only definition we need.

Then:

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

#color(white)(d/(dx) cos(arcsin(x))) = 1/2(1-x^2)^(-1/2)*(-2x)#

#color(white)(d/(dx) cos(arcsin(x))) = -x/sqrt(1-x^2)#

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Answer 2

William Abdalla

To find the derivative of ( \cos(\arcsin(x)) ), we use the chain rule. The derivative is ( \frac{-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.