How do you find the derivative of #y= x(1-x^2)^(1/2) + arccos(x)#?

Answer 1

Aurora Alvarado

The derivative is #dy/dx=-(2x^2)/sqrt(1-x^2)# for #-1 < x < 1#

Use the Product Rule and Chain Rule, along with the fact that #d/dx(arccos(x))=-1/sqrt{1-x^2}# (for #-1

#dy/dx=(1-x^2)^{1/2}+1/2 x(1-x^2)^{-1/2}*(-2x)-1/sqrt{1-x^2}#

Now simplify by getting a common denominator of #sqrt{1-x^2}#

#dy/dx=\frac{1-x^2-x^2-1}{sqrt{1-x^2}}=-(2x^2)/sqrt(1-x^2)#

for #-1 < x < 1#.

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Answer 2

Charles Arocha

To find the derivative of ( y = x(1 - x^2)^{1/2} + \arccos(x) ), apply the sum rule and the chain rule. The derivative of the first term involves both the product rule and the chain rule, while the derivative of the second term is straightforward.

Derivative of the first term: ( y_1 = x(1 - x^2)^{1/2} ) Apply the product rule: ( u = x ), ( v = (1 - x^2)^{1/2} ) ( u' = 1 ), ( v' = \frac{-x}{\sqrt{1 - x^2}} ) ( y_1' = u'v + uv' ) Substitute the values: ( y_1' = 1 \cdot (1 - x^2)^{1/2} + x \cdot \frac{-x}{\sqrt{1 - x^2}} )
Derivative of the second term: ( y_2 = \arccos(x) ) ( y_2' = -\frac{1}{\sqrt{1 - x^2}} )

Combine the derivatives of both terms: ( y' = y_1' + y_2' )

Final derivative: ( y' = (1 - x^2)^{1/2} - \frac{x^2}{\sqrt{1 - x^2}} - \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.