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How do you differentiate #y= ln(1-x^2)^(1/2)#?

Answer 1

Elijah Abram

#dy/dx=(-x)/(1-x^2)#

#color(green)(lna^b=blna)#

#y=ln(1-x^2)^(1/2)=1/2ln(1-x^2)#

#color(green)(d/dxlnu=1/u*color(blue)((du)/dx##rarrcolor(red)("Chain Rule")#

Differentiate,

#dy/dx=1/2*(1/(1-x^2))*color(blue)((0-2x)#

Simplify,

#dy/dx=(-x)/(1-x^2)#

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Answer 2

Cameron Alexander

To differentiate ( y = \ln\left((1 - x^2)^{\frac{1}{2}}\right) ), you can use the chain rule.

Let ( u = (1 - x^2)^{\frac{1}{2}} ). Then, differentiate ( u ) with respect to ( x ) to find ( \frac{du}{dx} ). Afterward, apply the chain rule to find ( \frac{dy}{dx} ).

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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.