Compute the gradient of the function at the given point.
-f(x, y) = ln(-5x + 3y), (-9, -4)
A)${\nabla} f = - \frac { 3 } { 11 } i - \frac { 4 } { 33 } j$
B) ${\nabla} f = - \frac { 5 } { 33 } \mathbf { i } + \frac { 1 } { 11 } j$
C)${\nabla} f = - \mathrm { i } + \frac { 1 } { 33 } \mathrm { j }$
D) ${\nabla} f = - \frac { 4 } { 33 } \mathrm { i } - \frac { 3 } { 11 } \mathrm { j }$

Question

Compute the gradient of the function at the given point.
-f(x, y) = ln(-5x + 3y), (-9, -4) 
A)${\nabla} f = - \frac { 3 } { 11 } i - \frac { 4 } { 33 } j$
B) ${\nabla} f = - \frac { 5 } { 33 } \mathbf { i } + \frac { 1 } { 11 } j$
C)${\nabla} f  = - \mathrm { i } + \frac { 1 } { 33 } \mathrm { j }$
D) ${\nabla} f = - \frac { 4 } { 33 } \mathrm { i } - \frac { 3 } { 11 } \mathrm { j }$

Quizplus · Accepted Answer

The gradient of $f(x, y) = \ln(-5x + 3y)$ is computed by taking the partial derivatives with respect to $x$ and $y$. For $x$, $\frac{\partial f}{\partial x} = \frac{-5}{-5x + 3y}$. For $y$, $\frac{\partial f}{\partial y} = \frac{3}{-5x + 3y}$. At the point $(-9, -4)$, these derivatives become $\frac{-5}{-5(-9) + 3(-4)} = \frac{-5}{45 - 12} = \frac{-5}{33}$ and $\frac{3}{33}$, respectively. Thus, the gradient is $\nabla f = - \frac{5}{33} \mathbf{i} + \frac{1}{11} \mathbf{j}$.

Compute the Gradient of the Function at the Given Point ∇f=−311i−433j{\nabla} f = - \frac { 3 } { 11 } i - \frac { 4 } { 33 } j∇f=−113​i−334​j

Compute the Gradient of the Function at the Given Point ${\nabla} f = - \frac { 3 } { 11 } i - \frac { 4 } { 33 } j$