The value of frac{d}{dx} tan^{-1} left[ frac{ | vedclass

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(B) Let $y = 	an^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} ight]$.Substitute $x = a 	an 	heta$,which implies $	heta = 	an^{-1}(\frac{x}{a})

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$\frac{3}{a}$

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(B) Let $y = 	an^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} ight]$.Substitute $x = a 	an 	heta$,which implies $	heta = 	an^{-1}(\frac{x}{a})$.Then,$y = 	an^{-1} \left[ \frac{3a^3 	an 	heta - a^3 	an^3 	heta}{a(a^2 - 3a^2 	an^2 	heta)} ight] = 	an^{-1} \left[ \frac{a^3(3 	an 	heta - 	an^3 	heta)}{a^3(1 - 3 	an^2 	heta)} ight]$.Using the trigonometric identity $	an(3	heta) = \frac{3 	an 	heta - 	an^3 	heta}{1 - 3 	an^2 	heta}$,we get $y = 	an^{-1}(	an 3	heta) = 3	heta$.Since $	heta = 	an^{-1}(\frac{x}{a})$,we have $y = 3 	an^{-1}(\frac{x}{a})$.Differentiating with respect to $x$:$\frac{dy}{dx} = 3 \cdot \frac{1}{1 + (x/a)^2} \cdot \frac{1}{a} = 3 \cdot \frac{a^2}{a^2 + x^2} \cdot \frac{1}{a} = \frac{3a}{a^2 + x^2}$.At $x = 0$,$\frac{dy}{dx} = \frac{3a}{a^2 + 0} = \frac{3}{a}$.

The value of $\frac{d}{dx} \tan^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} \right]$ at $x = 0$ is

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