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How do you find the derivative of the function #f(x)=arcsinx+arccosx#?

Answer 1

Notice the pattern between the two derivatives:

#d/(dx)[arcsinu] = 1/sqrt(1 - u^2)((du)/(dx))#

#d/(dx)[arccosu] = -1/sqrt(1 - u^2)((du)/(dx))#

Now you really only have to know one of them. #co# implies negative, then. It's also true for #d/(dx)[arc cotu]# vs. #d/(dx)[arc tanu]#, and #d/(dx)[arcsecu]# vs. #d/(dx)[arc cscu]#.

Now just do it.

#= 1/sqrt(1 - x^2) - 1/sqrt(1 - x^2) = 0#

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Answer 2

Ezra Adams

To find the derivative of ( f(x) = \arcsin(x) + \arccos(x) ), use the derivatives of ( \arcsin(x) ) and ( \arccos(x) ), which are ( \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} ) and ( \frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1-x^2}} ), respectively. Thus, the derivative of ( f(x) ) is ( f'(x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} ). Simplifying, we get ( f'(x) = 0 ).

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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.