Answer: (a) \(\ln x-1\). (b) \(\displaystyle y=\frac{x^3}{6}+x^2+x\ln x-2x+\mathrm{e}\).
Answer: (a) \(\ln x-1\). (b) \(\displaystyle y=\frac{x^3}{6}+x^2+x\ln x-2x+\mathrm{e}\).
(a) Differentiate using the product rule:
\(\displaystyle \frac{d}{dx}(x\ln x)=\ln x+1\).
Also,
\(\displaystyle \frac{d}{dx}(2x)=2\).
Therefore
\(\displaystyle \frac{d}{dx}(x\ln x-2x)=\ln x+1-2=\ln x-1\).
(b) First simplify the second derivative:
\(\displaystyle \left(\frac{x+1}{\sqrt{x}}\right)^2=\frac{(x+1)^2}{x}=x+2+\frac1x.\)
So
\(\displaystyle \frac{d^2y}{dx^2}=x+2+\frac1x\).
Integrate once:
\(\displaystyle \frac{dy}{dx}=\frac{x^2}{2}+2x+\ln x+C\).
At \(x=\mathrm{e}\), the gradient is \(\frac{\mathrm{e}^2}{2}+2\mathrm{e}\). Since \(\ln \mathrm{e}=1\),
\(\displaystyle \frac{\mathrm{e}^2}{2}+2\mathrm{e}+1+C=\frac{\mathrm{e}^2}{2}+2\mathrm{e}\).
Hence \(C=-1\), so
\(\displaystyle \frac{dy}{dx}=\frac{x^2}{2}+2x+\ln x-1\).
Now integrate again:
\(\displaystyle y=\frac{x^3}{6}+x^2+\int(\ln x-1)\,dx+C_2\).
From part (a), an antiderivative of \(\ln x-1\) is \(x\ln x-2x\). Therefore
\(\displaystyle y=\frac{x^3}{6}+x^2+x\ln x-2x+C_2\).
The curve passes through \(\left(\mathrm{e},\frac{\mathrm{e}^3}{6}+\mathrm{e}^2\right)\), so
\(\displaystyle \frac{\mathrm{e}^3}{6}+\mathrm{e}^2=\frac{\mathrm{e}^3}{6}+\mathrm{e}^2+\mathrm{e}\ln\mathrm{e}-2\mathrm{e}+C_2.\)
Since \(\ln\mathrm{e}=1\), this becomes
\(\displaystyle \frac{\mathrm{e}^3}{6}+\mathrm{e}^2=\frac{\mathrm{e}^3}{6}+\mathrm{e}^2-\mathrm{e}+C_2.\)
Thus \(C_2=\mathrm{e}\), giving
\(\displaystyle y=\frac{x^3}{6}+x^2+x\ln x-2x+\mathrm{e}\).
