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

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Answer 1 · 2015-06-17T13:17:14

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 $-1<x<1$ ).

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

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 $-1 < x < 1$ .

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