यदि y=x log left(frac{1}{a x}+frac{1}{a}right | vedclass

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(C) दिया गया है $y=x \log \left(\frac{1}{a x}+\frac{1}{a}\right) = x \log \left(\frac{1+x}{ax}\right)$.सबसे पहले,प्रथम अवकलज $\frac{dy}{dx}$ ज्ञात करे

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$-1$

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(C) दिया गया है $y=x \log \left(\frac{1}{a x}+\frac{1}{a}\right) = x \log \left(\frac{1+x}{ax}\right)$.सबसे पहले,प्रथम अवकलज $\frac{dy}{dx}$ ज्ञात करें:$\frac{dy}{dx} = \log \left(\frac{1+x}{ax}\right) + x \cdot \frac{ax}{1+x} \cdot \frac{d}{dx} \left(\frac{1+x}{ax}\right)$$= \log \left(\frac{1+x}{ax}\right) + \frac{ax^2}{1+x} \cdot \left(\frac{ax - a(1+x)}{(ax)^2}\right)$$= \log \left(\frac{1+x}{ax}\right) + \frac{ax^2}{1+x} \cdot \left(\frac{-a}{a^2x^2}\right) = \log \left(\frac{1+x}{ax}\right) - \frac{1}{1+x}$.अब,द्वितीय अवकलज $\frac{d^2y}{dx^2}$ ज्ञात करें:$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\log \left(\frac{1+x}{ax}\right) - \frac{1}{1+x}\right)$$= \frac{ax}{1+x} \cdot \left(\frac{-1}{ax^2}\right) + \frac{1}{(1+x)^2} = -\frac{1}{x(1+x)} + \frac{1}{(1+x)^2} = \frac{-(1+x) + x}{x(1+x)^2} = \frac{-1}{x(1+x)^2}$.इन मानों को $x(x+1) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y$ में प्रतिस्थापित करें:$= x(x+1) \left(\frac{-1}{x(1+x)^2}\right) + x \left(\log \left(\frac{1+x}{ax}\right) - \frac{1}{1+x}\right) - x \log \left(\frac{1+x}{ax}\right)$$= -\frac{1}{1+x} + x \log \left(\frac{1+x}{ax}\right) - \frac{x}{1+x} - x \log \left(\frac{1+x}{ax}\right)$$= -\left(\frac{1+x}{1+x}\right) = -1$.

यदि $y=x \log \left(\frac{1}{a x}+\frac{1}{a}\right)$ है,तो $x(x+1) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=$

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