If \( y=x\log_{e}\left(\displaystyle\frac{x}{2+x}\right) \), then \( x^{3}\displaystyle\frac{d^{2}y}{dx^{2}} \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \left(\displaystyle\frac{2x}{2+x}\right)^{2} \)
Splitting the logarithm,
\( y=x\log_{e}x-x\log_{e}(2+x) \)
\( \displaystyle \frac{dy}{dx}=\log_{e}x+1-\log_{e}(2+x)-\frac{x}{2+x} \)
Since \( \frac{x}{2+x}=1-\frac{2}{2+x} \), this simplifies to
\( \displaystyle \frac{dy}{dx}=\log_{e}x-\log_{e}(2+x)+\frac{2}{2+x} \)
Differentiating again:
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{1}{x}-\frac{1}{2+x}-\frac{2}{(2+x)^{2}} \)
\( \displaystyle =\frac{2}{x(2+x)}-\frac{2}{(2+x)^{2}} \)
\( \displaystyle =\frac{2(2+x)-2x}{x(2+x)^{2}} \)
\( \displaystyle =\frac{4}{x(2+x)^{2}} \)
Multiplying by \( x^{3} \):
\( \displaystyle x^{3}\frac{d^{2}y}{dx^{2}}=\frac{4x^{2}}{(2+x)^{2}} \)
\( \displaystyle =\left(\frac{2x}{2+x}\right)^{2} \)