Find $f _ { y } ( - 18,6 )$ for $f ( x , y ) = \sin ( 3 x + 9 y )$.
Select the correct answer.
A) 0
B) 9
C) $- 9$
D) $- 3$
E) 3

Question

Quizplus · Accepted Answer

To find the partial derivative of $f$ with respect to $y$, treating $x$ as a constant, we simply differentiate the function with respect to $y$:

$$f_y(x,y) = 9\cos(3x+9y)$$

Plugging in $x=-18$ and $y=6$ yields $f_y(-18,6) = 9\cos(-162) = 9\cos(18)$. Since $\cos(18)$ is positive, the answer is $\boxed{	extbf{(B)}\ 9}$.

Find fy(−18,6)f _ { y } ( - 18,6 )fy​(−18,6) For f(x,y)=sin⁡(3x+9y)f ( x , y ) = \sin ( 3 x + 9 y )f(x,y)=sin(3x+9y)