"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)."

I don't know how to solve this problem. Could someone walk me through it? Thank you so much!

a) find the partial derivatives of each, and just plug in 1 for x and 3 for y (values should be close to 0)/

b) in using the difference quotients, take the limit as h approaches infinity of both each derivative, plugging in 0.1 for x and y in each one. It's better that you think critically about how the definition of the derivatives applies not only in 2D, but in 3-space as well. You'll gain a better intuition.

Of course! To estimate fx(1,3) and fy(1,3) using difference quotients, we can use the formulas:

fx(x, y) ≈ (f(x + delta(x), y) - f(x, y)) / delta(x)

fy(x, y) ≈ (f(x, y + delta(y)) - f(x, y)) / delta(y)

where delta(x) and delta(y) are the given step sizes.

Given that delta(x) = 0.1 and delta(y) = 0.1, we can apply these formulas to estimate the partial derivatives of f(x, y) = e^(-x)sin(y) at the point (1, 3).

First, let's estimate fx(1, 3):

fx(1, 3) ≈ (f(1 + 0.1, 3) - f(1, 3)) / 0.1

To calculate f(1 + 0.1, 3), substitute x = 1.1 and y = 3 into f(x, y) = e^(-x)sin(y):

f(1.1, 3) = e^(-1.1)sin(3)

Next, substitute x = 1 and y = 3 into f(x, y) = e^(-x)sin(y) to find f(1, 3).

fx(1, 3) ≈ (f(1.1, 3) - f(1, 3)) / 0.1

Now, let's estimate fy(1, 3):

fy(1, 3) ≈ (f(1, 3 + 0.1) - f(1, 3)) / 0.1

Substitute x = 1 and y = 3.1 into f(x, y) = e^(-x)sin(y) to find f(1, 3.1).

Lastly, substitute x = 1 and y = 3 into f(x, y) = e^(-x)sin(y) to find f(1, 3).

fy(1, 3) ≈ (f(1, 3 + 0.1) - f(1, 3)) / 0.1

By substituting the appropriate values and evaluating the expressions, you can estimate fx(1, 3) and fy(1, 3).