Find $\nabla f$ for $f ( x , y ) = \frac { \tan ( x ) } { y }$ . A) $$ B) $$ C) $$ D) $$

Question

Find $\nabla f$ for $f ( x , y ) = \frac { \tan ( x ) } { y }$ . A) $< - \frac { \tan ( x ) } { y ^ { 2 } } , \frac { \sec ^ { 2 } ( x ) } { y ^ { 2 } } >$ B) $< \frac { \tan ( x ) } { y } , - \frac { \tan ( x ) } { y ^ { 2 } } >$ C) $< \frac { \sec ^ { 2 } ( x ) } { y } , - \frac { \tan ( x ) } { y ^ { 2 } } >$ D) $< \frac { \sec ^ { 2 } ( x ) } { y ^ { 2 } } , \frac { \tan ( x ) } { y } >$

Quizplus · Accepted Answer

The gradient $\nabla f$ of a function $f(x, y)$ is given by the vector of its partial derivatives, $\nabla f = \left< \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right>$. For $f(x, y) = \frac{\tan(x)}{y}$, the partial derivative with respect to $x$ is $\frac{\sec^2(x)}{y}$ because the derivative of $\tan(x)$ with respect to $x$ is $\sec^2(x)$, and $y$ is treated as a constant. The partial derivative with respect to $y$ is $-\frac{\tan(x)}{y^2}$ because the derivative of $\frac{1}{y}$ with respect to $y$ is $-\frac{1}{y^2}$, and $\tan(x)$ is treated as a constant.

Find ∇f\nabla f∇f For f(x,y)=tan⁡(x)yf ( x , y ) = \frac { \tan ( x ) } { y }f(x,y)=ytan(x)​

Find $\nabla f$ For $f ( x , y ) = \frac { \tan ( x ) } { y }$