If $y=\left(\frac{x^{2}}{x+1}\right)^{x}$ and $\frac{d y}{d x}=y\left[g(x)+\log \left(\frac{x^{2}}{x+1}\right)\right]$, then $g(x)=$

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Aicharya · Accepted Answer

1. Express $y$ in terms of exponentials: $y=\left(\frac{x^2}{x+1}\right)^x=e^{x \log \left(\frac{x^2}{x+1}\right)}$ 2. Differentiate $y$ using the chain rule: Taking the natural logarithm of both sides: $\ln (y)=x \log \left(\frac{x^2}{x+1}\right)$ Differentiating both sides with respect to $x$ : $\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}\left[x \log \left(\frac{x^2}{x+1}\right)\right] .$ 3. Differentiate $x \log \left(\frac{x^2}{x+1}\right)$ : Using the product rule: $\frac{d}{d x}\left[x \log \left(\frac{x^2}{x+1}\right)\right]=\log \left(\frac{x^2}{x+1}\right)+x \cdot \frac{d}{d x} \log \left(\frac{x^2}{x+1}\right)$ Simplify: $\frac{d}{d x}\left[x \log \left(\frac{x^2}{x+1}\right)\right]=\log \left(\frac{x^2}{x+1}\right)+\frac{x+2}{x+1}$ 6. Relate to the given equation: Substituting $\frac{1}{y} \frac{d y}{d x}$ : $\frac{1}{y} \frac{d y}{d x}=\log \left(\frac{x^2}{x+1}\right)+\frac{x+2}{x+1}$ Comparing this with the given form: $\frac{1}{y} \frac{d y}{d x}=g(x)+\log \left(\frac{x^2}{x+1}\right)$ we find: $g(x)=\frac{x+2}{x+1}$

If $y=\left(\frac{x^{2}}{x+1}\right)^{x}$ and $\frac{d y}{d x}=y\left[g(x)+\log…

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