Question

Let \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},\) and let \(f(x, y, z)=\|\mathbf{F}(x, y, z)\| .\) $$ \text { Show that } \nabla(\ln f)=\frac{\mathbf{F}}{f^{2}} $$

Short answer

After calculating gradients and using chain rule for differentiation, we have shown that \(\nabla (\ln f) = \frac{F}{f^2}\). This demonstrates that the original statement given in the problem is correct.

Step-by-step solution

Compute gradient of the function

Begin the problem by calculating the gradient of \( f \). Here \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \) which is the magnitude of the vector function \( F \). So, the computation of its gradient will be:\n\(\nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f}{\partial z} \mathbf{k}\).\nUpon calculating, we get: \(\nabla f = \frac{x}{f} \mathbf{i} + \frac{y}{f} \mathbf{j} + \frac{z}{f} \mathbf{k} = \frac{F}{f} \)

Compute the required expression

Next, compute the gradient of \( \ln(f) \). Using the chain rule, we have:\n\(\nabla (\ln f) = \frac{1}{f}\nabla f\). Subsitute \(\nabla f\) from step 1, we obtain: \(\nabla (\ln f) = \frac{1}{f} \cdot \frac{F}{f} = \frac{F}{f^2}\)