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Calculate $frac{partial f}{partial x}, frac{partial f}{partial y},left.frac{partial f}{partial x}ight|_{(1,-1)}$, and $left.frac{partial f}{partial y}ight|_{(1,-1)}...

Question

Calculate $frac{partial f}{partial x}, frac{partial f}{partial y},left.frac{partial f}{partial x}ight|_{(1,-1)}$, and $left.frac{partial f}{partial y}ight|_{(1,-1)}$ when defined. HINT [See Quick Examples page 1098.]$$f(x, y)=x^{4} y^{2}-x$$

Calculate $frac{partial f}{partial x}, frac{partial f}{partial y},left.frac{partial f}{partial x} ight|_{(1,-1)}$, and $left.frac{partial f}{partial y} ight|_{(1,-1)}$ when defined. HINT [See Quick Examples page 1098.] $$ f(x, y)=x^{4} y^{2}-x $$



Answers

Calculate $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\left.\frac{\partial f}{\partial x}\right|_{(1,-1)}$, and $\left.\frac{\partial f}{\partial y}\right|_{(1,-1)}$ when defined. HINT [See Quick Examples page 1098.] $$ f(x, y)=x^{4} y^{2}-x $$

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F of X Y is equal to X squared minus y squared. So let's go ahead and calculate the derivative with respect to X. Get two X now the partial derivative with respect to why remember, we're treating ex like a constant this time, so we have negative two y.

We have of X Y is equal to one over X plus y. Let's go ahead and take the derivative with respect to X. So let's actually take a second and rewrite f. You're gonna rewrite it like with a negative one exponents. Then we have the negative one coming down X plus y the negative, too. Now, let's go ahead and figure out the partial derivative with respect. Why? And we actually get same thing.

We have f of X y is equal to e to the X, divided by one plus you to the why. Now, let's go ahead and take the derivative with respect to X. So remember that the derivative of each of the axes each the X that this other thing is treated as a constant. Now let's calculate the derivative with respect to why so but that let's go ahead and rewrite F one plus e to the y. So a negative one, then we have a negative one coming down, and this exponents changes to a negative, too.


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