# Thread: Difference quotients

1. ## Difference quotients

Here's the question:

"Use difference quotients with delta(x)=0.1 and delta(y)=0.1 to estimate fx(1,3) and fy(1,3) where f(x,y)=e^(-x)sin(y)."

My work so far:

I know that to begin with, you have to find the partial derivative with respect to both x and y and then plug the point (1,3) into this equation. So for x the derivative would be -1e^(-x)sin(y) and for y the derivative would be e^(-x)cos(y). Now, what do I do after I plug the point (1,3) into these two equations? Any help would be appreciated.

2. Here is how the difference quotient is used to define the first partial derivatives of f(x,y) at (1,3).

$f_x(1,3) = \lim_{\Delta x \to 0}\frac{f(1 + \Delta x,3) - f(1,3)}{\Delta x} \approx \frac{f(1 + 0.1,3) - f(1,3)}{0.1} = \frac{e^{-(1 + 0.1)}\sin{3} - e^{-1}\sin{3}}{0.1} = \frac{e^{-1.01}\sin{3} - e^{-1}\sin{3}}{0.1}$

$f_y(1,3) = \lim_{\Delta y \to 0}\frac{f(1,3 + \Delta y) - f(1,3)}{\Delta y} \approx \frac{f(1,3 + 0.1) - f(1,3)}{0.1} = \frac{e^{-1}\sin{3 + 0.1} - e^{-1}\sin{3}}{0.1} = \frac{e^{-1}\sin{3.1} - e^{-1}\sin{3}}{0.1}$