Question

Calculate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ then evaluate at (1, -1). (If an answer is undefined, enter UNDEFINED.) $\qquad f(x, y) = \frac{7}{(xy + 1)^1}$ $\frac{\partial f}{\partial x} = \text{ } $\frac{\partial f}{\partial y} = \text{ } $\left.\frac{\partial f}{\partial x}\right|_{(1, -1)} = \text{ } $\left.\frac{\partial f}{\partial y}\right|_{(1, -1)} = \text{ }

          Calculate $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ then evaluate at (1, -1). (If an answer is undefined, enter UNDEFINED.)
$\qquad f(x, y) = \frac{7}{(xy + 1)^1}$
$\frac{\partial f}{\partial x} = \text{ }
$\frac{\partial f}{\partial y} = \text{ }
$\left.\frac{\partial f}{\partial x}\right|_{(1, -1)} = \text{ }
$\left.\frac{\partial f}{\partial y}\right|_{(1, -1)} = \text{ }

Calculate (∂ f)/(∂ x) and (∂ f)/(∂ y) then evaluate at (1, -1). (If an answer is undefined, enter UNDEFINED.)
f(x, y) = (7)/((xy + 1)^1)
(∂ f)/(∂ x) =(∂f)/(∂y) =
.(∂ f)/(∂ x)|(1, -1) =.(∂f)/(∂y)|(1, -1) =

Added by Riley H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Gregory Higby

09/03/2022

Step 1

To find ax, we differentiate f(x,y) = (xy + 1)^1 with respect to x, treating y as a constant. Using the chain rule, we have: ax = d/dx [(xy + 1)^1] = 1 * (xy + 1)^(1-1) * d/dx (xy + 1) = (xy + 1)^0 * (y + 0) = y So, ax = y. Show more…

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Calculate and then evaluate at (1, -1). (If an answer is undefined, enter UNDEFINED.) af Calculate and af then evaluate at (1, -1). (If an answer is undefined, enter UNDEFINED.) ax ay 7 f(x,y) = (xy + 1)1 af ax af ay of ax(1,-1 af ay(1,-1)