find the gradient of the function f(x,y,z)=x^3ln(yz), at the point (-1,1,e) ∇f(-1,1,e)=<3, - 1/e, - 1/e>…

Explanation:

Step1: Recall gradient formula

The gradient of a function $f(x,y,z)$ is $\nabla f=\left\langle\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle$.

Step2: Calculate $\frac{\partial f}{\partial x}$

Given $f(x,y,z)=x^{3}\ln(yz)$, using the product - rule and chain - rule, $\frac{\partial f}{\partial x}=3x^{2}\ln(yz)$. Evaluate at $x = - 1,y = 1,z = e$: $\frac{\partial f}{\partial x}\big|_{(-1,1,e)}=3(-1)^{2}\ln(1\times e)=3$.

Step3: Calculate $\frac{\partial f}{\partial y}$

Using the product - rule and chain - rule, $\frac{\partial f}{\partial y}=\frac{x^{3}}{y}$. Evaluate at $x=-1,y = 1,z = e$: $\frac{\partial f}{\partial y}\big|_{(-1,1,e)}=\frac{(-1)^{3}}{1}=-1$.

Step4: Calculate $\frac{\partial f}{\partial z}$

Using the product - rule and chain - rule, $\frac{\partial f}{\partial z}=\frac{x^{3}}{z}$. Evaluate at $x=-1,y = 1,z = e$: $\frac{\partial f}{\partial z}\big|_{(-1,1,e)}=\frac{(-1)^{3}}{e}=-\frac{1}{e}$.

Answer:

$\left\langle3,-1,-\frac{1}{e}\right\rangle$