If y = tan^{-1} left( frac{x^{1/3} + a^{1/3}} | vedclass

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Question

(A) Given $y = 	an^{-1} \left( \frac{x^{1/3} + a^{1/3}}{1 - x^{1/3}a^{1/3}} ight)$.Using the formula $	an^{-1} \left( \frac{u + v}{1 - uv} ight)

Vedclass · Accepted Answer

$\frac{1}{3x^{2/3}(1 + x^{2/3})}$

Vedclass · Answer

(A) Given $y = 	an^{-1} \left( \frac{x^{1/3} + a^{1/3}}{1 - x^{1/3}a^{1/3}} ight)$.Using the formula $	an^{-1} \left( \frac{u + v}{1 - uv} ight) = 	an^{-1} u + 	an^{-1} v$,we get:$y = 	an^{-1}(x^{1/3}) + 	an^{-1}(a^{1/3})$.Differentiating both sides with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [	an^{-1}(x^{1/3})] + \frac{d}{dx} [	an^{-1}(a^{1/3})]$.Since $a$ is a constant,$\frac{d}{dx} [	an^{-1}(a^{1/3})] = 0$.Using the chain rule,$\frac{d}{dx} [	an^{-1}(x^{1/3})] = \frac{1}{1 + (x^{1/3})^2} \cdot \frac{d}{dx}(x^{1/3})$.$= \frac{1}{1 + x^{2/3}} \cdot \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}(1 + x^{2/3})}$.Thus,$\frac{dy}{dx} = \frac{1}{3x^{2/3}(1 + x^{2/3})}$.

If $y = \tan^{-1} \left( \frac{x^{1/3} + a^{1/3}}{1 - x^{1/3}a^{1/3}} \right)$,then $\frac{dy}{dx} = $

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