If y=x log left(frac{x}{2-3 x}right) for 0

Question

(C) Given,$y=x \log \left(\frac{x}{2-3 x}\right)$.Applying the product rule and chain rule to differentiate with respect to $x$:$\frac{dy}{dx} = x \cd

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Given,$y=x \log \left(\frac{x}{2-3 x}\right)$.Applying the product rule and chain rule to differentiate with respect to $x$:$\frac{dy}{dx} = x \cdot \frac{d}{dx} \left[ \log \left( \frac{x}{2-3x} \right) \right] + \log \left( \frac{x}{2-3x} \right) \cdot \frac{d}{dx}(x)$$\frac{dy}{dx} = x \left( \frac{2-3x}{x} \right) \cdot \frac{(2-3x)(1) - x(-3)}{(2-3x)^2} + \log \left( \frac{x}{2-3x} \right)$$\frac{dy}{dx} = x \left( \frac{2-3x}{x} \right) \cdot \frac{2}{(2-3x)^2} + \log \left( \frac{x}{2-3x} \right) = \frac{2}{2-3x} + \log \left( \frac{x}{2-3x} \right)$.Now,differentiate again with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( 2(2-3x)^{-1} \right) + \frac{d}{dx} \left[ \log \left( \frac{x}{2-3x} \right) \right]$$\frac{d^2y}{dx^2} = -2(2-3x)^{-2}(-3) + \frac{2}{x(2-3x)} = \frac{6}{(2-3x)^2} + \frac{2}{x(2-3x)}$.At $x = \frac{1}{2}$,$2-3x = 2 - 3(\frac{1}{2}) = 2 - 1.5 = 0.5 = \frac{1}{2}$.Substituting $x = \frac{1}{2}$ and $2-3x = \frac{1}{2}$:$\frac{d^2y}{dx^2} = \frac{6}{(1/2)^2} + \frac{2}{(1/2)(1/2)} = \frac{6}{1/4} + \frac{2}{1/4} = 24 + 8 = 32$.

If $y=x \log \left(\frac{x}{2-3 x}\right)$ for $0 < x < \frac{2}{3}$,then $\frac{d^2 y}{d x^2}$ at $x=\frac{1}{2}$ is

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