Q.If $w(x,y,z)=x^2+y^2+z^2,\ x=e^t,\ y=e^t\sin t$ and $z=e^t\cos t$, find $\dfrac{dw}{dt}$.

Since $x$ and $y$ are each functions of the single variable $t$, the total derivative chain rule uses ordinary derivatives $dx/dt$ and $dy/dt$ (not partial derivatives $\partial x/\partial t$), matching option (b)/(c).

1. $u=f(x,y)$ is a differentiable function of two variables $x,y$, and both $x=x(t)$, $y=y(t)$ are differentiable functions of the single independent variable $t$.
2. Because $x$ and $y$ each depend on $t$ alone (not on $t$ together with any other independent variable), the rate of change of $x$ and $y$ with respect to $t$ is an ordinary derivative, written $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ — not a partial derivative $\dfrac{\partial x}{\partial t}$, which would only be meaningful if $x$ depended on $t$ and some other independent variable simultaneously.
3. The chain rule for the total derivative of the composite $u=f(x(t),y(t))$ is therefore: $\frac{du}{dt} = \frac{\partial f}{\partial x}\cdot\frac{dx}{dt} + \frac{\partial f}{\partial y}\cdot\frac{dy}{dt}$
4. Options (a) and (d) incorrectly use $\dfrac{\partial x}{\partial t}$ and $\dfrac{\partial y}{\partial t}$ in place of the ordinary derivatives, and (d) also incorrectly writes $\dfrac{\partial u}{\partial t}$ on the left (there is no such partial derivative here, since $u$'s only route of dependence on $t$ is through $x$ and $y$) — both are wrong.
5. Options (b) and (c) both correctly state $\dfrac{du}{dt}=\dfrac{\partial f}{\partial x}\cdot\dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\cdot\dfrac{dy}{dt}$.
6. Source-transcription note: in the original question paper's transcription used here, options (b) and (c) are printed as literally identical text — this is a duplication artifact in the source, not a genuine second correct-looking alternative to weigh against (a)/(d). The mathematics is unambiguous; either letter (b) or (c) represents the one correct formula.