How do you use chain rule with a product rule to differentiate #y = x*sqrt(1-x^2)#?

Answer 1

If we selected a number for #x# and did the arithmetic, then the last operation would be multiplication. So, ultimately this is a product.

The derivative w.r.t. #x# of the first function is #1#.

To find the derivative of the second function, we'll need the power rule (the derivative of the square root) and the chain rule. (Because we're not just taking square root of #x#.)

Using the product rule in the form: #(FS)' = F'S+FS'# we get:

#y' = (1)(sqrt(1-x^2)) + (x)(1/(2sqrt(1-x^2))*(-2x))#

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

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