Inverse trig derivatives (1 Viewer)

Bohimei · Aug 15, 2016

Differentiate y= cos^(-1)(1-x^2)/(1+x^2)

Thanks

InteGrand · Aug 15, 2016

Bohimei said:
Differentiate y= cos^(-1)(1-x^2)/(1+x^2)

Thanks

$\noindent Let $y = \cos ^{-1} \left( \frac{1-x^2}{1+x^2}\right)$. To find $y^\prime$, use the chain rule. The derivative of the inner function is $\frac{-2x \left(1+x^2\right) -\left(1-x^2\right)\cdot 2x}{\left(1+x^2\right)^2} = \frac{-4x}{\left(1+x^2\right)^2}$. So $y^\prime = \frac{-4x}{\left(1+x^2\right)^2} \times \frac{-1}{\sqrt{1- \left( \frac{1-x^2}{1+x^2} \right)^2}}$. The rest is an exercise in algebraic simplification.$

Inverse trig derivatives (1 Viewer)

Bohimei

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InteGrand

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