What is the derivative of this function #y=x^2+cos^-1x#?

Answer 1

Christopher Allers

#y' = 2x - 1/sinx#

#y = x^2 + cos^-1 x# #y = x^2 + 1/cosx# #d/dx y = d/dx (x^2 + 1/cosx)# #y' = 2x + (1/-sinx)# #y' = 2x - 1/sinx#

you have to remember the trig table of derivatives #d/dx sinx = cosx# #d/dx cosx = -sinx#

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Answer 2

Isabella Ackerman

To find the derivative of the function ( y = x^2 + \cos^{-1} x ), you can apply the sum rule and the chain rule. The derivative is:

[ y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(\cos^{-1} x) ]

[ y' = 2x - \frac{1}{\sqrt{1-x^2}} ]

So, the derivative of the function is ( y' = 2x - \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.