Find the directional derivative of the function $f ( x , y ) = y ^ { 2 } \ln x$ at the point $( 1,2 )$ in the direction $( 3,4 )$ .
A) $\frac { 5 } { 16 }$

B)0
C) $\frac { 12 } { 5 }$
D)12
E)1
F) $\frac { 16 } { 5 }$

G) $\frac { 5 } { 12 }$
H) $\frac { 1 } { 12 }$

Question

Find the directional derivative of the function $f ( x , y ) = y ^ { 2 } \ln x$ at the point $( 1,2 )$ in the direction $( 3,4 )$ .
A) $\frac { 5 } { 16 }$

B)0
C) $\frac { 12 } { 5 }$ 
D)12
E)1
F) $\frac { 16 } { 5 }$

G) $\frac { 5 } { 12 }$ 
H) $\frac { 1 } { 12 }$

Quizplus · Accepted Answer

The directional derivative of a function $f(x, y)$ at a point $(x_0, y_0)$ in the direction of a vector $\mathbf{u} = (a, b)$, where $\mathbf{u}$ is a unit vector, is given by $\nabla f(x_0, y_0) \cdot \mathbf{u}$. First, we find the gradient of $f$, $\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$. For $f(x, y) = y^2 \ln x$, we have $\frac{\partial f}{\partial x} = \frac{y^2}{x}$ and $\frac{\partial f}{\partial y} = 2y \ln x$. At $(1, 2)$, $\nabla f = (4, 4 \ln 1) = (4, 0)$. The direction vector $(3, 4)$ must be normalized: $\mathbf{u} = \frac{(3, 4)}{\sqrt{3^2 + 4^2}} = \frac{(3, 4)}{5}$. The directional derivative is $\nabla f \cdot \mathbf{u} = (4, 0) \cdot \left(\frac{3}{5}, \frac{4}{5}\right) = \frac{12}{5}$.

Find the Directional Derivative of the Function f(x,y)=y2ln⁡xf ( x , y ) = y ^ { 2 } \ln xf(x,y)=y2lnx

Find the Directional Derivative of the Function $f ( x , y ) = y ^ { 2 } \ln x$