If $u=x^2+y^2$ and $x=s+3t, y=2s-t,$ then $\frac{d^{2}u}{d{s}^2}$ is equal to (where $t$ is a constant)

Given, $u=x^2+y^2,x=s+3t, y=2s-t$

Now ,$\frac{dx}{ds}=1,\frac{dy}{ds}=2\dots \left(\mathrm{i}\right)$

$\frac{d^{2}x}{d{s}^2}=0,\frac{d^{2}y}{d{s}^2}=0\dots \left(\mathrm{i}\mathrm{i}\right)$

Now, $u=x^2+y^2$

$\frac{du}{ds}=2x\frac{dx}{ds}+2y\frac{dy}{ds}$

$\frac{d^{2}u}{d{s}^2}=2{\left(\frac{dx}{ds}\right)}^2+2x\frac{d^{2}x}{d{s}^2}+2{\left(\frac{dy}{ds}\right)}^2+2y\left(\frac{d^{2}y}{d{s}^2}\right)$

$\Rightarrow \frac{d^{2}u}{d{s}^2}=2{\left(1\right)}^2+2x\left(0\right)+2{\left(2\right)}^2+2y\left(0\right) (\mathrm{from} (i) \mathrm{and} (ii\left)\right)$

$=2+8=10$