The derivative of frac{1-x^2}{1+x^2} with res | vedclass

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(B) Let $u = \frac{1-x^2}{1+x^2}$ and $v = \frac{2x}{1+x^2}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.First, find $\frac{du}{dx}$:$\frac{

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$\frac{4}{3}$

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(B) Let $u = \frac{1-x^2}{1+x^2}$ and $v = \frac{2x}{1+x^2}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.First, find $\frac{du}{dx}$:$\frac{du}{dx} = \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} = \frac{-4x}{(1+x^2)^2}$.Next, find $\frac{dv}{dx}$:$\frac{dv}{dx} = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2}$.Now, calculate $\frac{du}{dv}$:$\frac{du}{dv} = \frac{-4x}{(1+x^2)^2} \div \frac{2(1-x^2)}{(1+x^2)^2} = \frac{-4x}{2(1-x^2)} = \frac{-2x}{1-x^2} = \frac{2x}{x^2-1}$.At $x=2$:$\frac{du}{dv} = \frac{2(2)}{2^2-1} = \frac{4}{4-1} = \frac{4}{3}$.

The derivative of $\frac{1-x^2}{1+x^2}$ with respect to $\frac{2x}{1+x^2}$ at $x=2$ is

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