Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 17

Treat $R$ and $a$ as constants and use the total differential in the independent variables $n,V,T$:Each coefficient differentiates with respect to its displayed variable while holding the other two fixed; the formula applies where $V\ne 0$.

For $r=\sqrt{x^2+y^2+z^2}>0$, the multivariable chain rule gives ${\partial }_{x}g(r)=g^{\prime }(r)x/r$, with the corresponding expressions for $y,z$. Thus the gradient of a radial function isThe origin needs a separate check because $\hat{\mathbf{r}}$ is undefined there. If the one-variable function has a finite right derivative at zero, the radial composite is differentiable at the origin exactly when $g^{\prime }(0^{+})=0$, in which case its gradient is zero. A nonzero radial slope gives incompatible opposite-direction derivatives.

The gradient is $\nabla f=(2x,z,y)$, so its value at the point is $(2,-3,1)$. Since the specified direction is already a unit vector, the directional derivative is its dot product with that gradient:

For a unit direction $\mathbf{v}$, the Cauchy-Schwarz inequality gives $\nabla f\cdot \mathbf{v}\le |\nabla f|$, with equality in the gradient direction. Its norm here is $\sqrt{4+9+1}=\sqrt{14}$, so the direction of fastest increase isThe maximal directional derivative itself is $\sqrt{14}$.