Answer: \(\displaystyle y=\frac{4}{15}x^{5/2}+x^2+\frac43x^{3/2}-\frac{10}{3}x-\frac{4}{15}\).
Answer: \(\displaystyle y=\frac{4}{15}x^{5/2}+x^2+\frac43x^{3/2}-\frac{10}{3}x-\frac{4}{15}\).
First simplify the second derivative:
\(\displaystyle \left(\frac{\sqrt{x}+1}{\sqrt[4]{x}}\right)^2=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}\).
Expanding the numerator gives
\(x+2\sqrt{x}+1\).
Therefore
\(\displaystyle \frac{d^2y}{dx^2}=x^{1/2}+2+x^{-1/2}\).
Integrate once:
\(\displaystyle \frac{dy}{dx}=\frac23x^{3/2}+2x+2x^{1/2}+C\).
The gradient is \(\frac43\) at \(x=1\), so
\(\displaystyle \frac43=\frac23+2+2+C\).
Thus
\(\displaystyle C=-\frac{10}{3}\).
So
\(\displaystyle \frac{dy}{dx}=\frac23x^{3/2}+2x+2x^{1/2}-\frac{10}{3}.\)
Integrate again:
\(\displaystyle y=\frac{4}{15}x^{5/2}+x^2+\frac43x^{3/2}-\frac{10}{3}x+D.\)
The curve passes through \((1,-1)\), so
\(\displaystyle -1=\frac{4}{15}+1+rac43-rac{10}{3}+D\).
The numerical part is
\(\displaystyle \frac{4}{15}+\frac{15}{15}+\frac{20}{15}-\frac{50}{15}=-\frac{11}{15}.\)
Hence
\(\displaystyle -1=-\frac{11}{15}+D\),
so
\(\displaystyle D=-\frac{4}{15}\).
Therefore the equation of the curve is
\(\displaystyle y=\frac{4}{15}x^{5/2}+x^2+\frac43x^{3/2}-\frac{10}{3}x-\frac{4}{15}.\)
