Suppose that z = e x 2 y where x = ? u v and y = 1 v . Find partial z/ partial u and ...


Suppose that {eq}z = e^{x^2y} {/eq} where {eq}x = \sqrt{uv} {/eq} and {eq}y=\frac1v {/eq}. Find {eq}\frac{\partial z}{\partial u} {/eq} and {eq}\frac{\partial z}{\partial v} {/eq} in terms of {eq}u {/eq} and {eq}v {/eq}. Simplify your answers

Evaluate First Partial Derivatives:

Suppose a function, {eq}f(x,y) {/eq}, where {eq}x=g(u,v) \\ y=h(u,v) {/eq}

Then the first partial derivatives of {eq}f {/eq} with respect to {eq}u {/eq} and {eq}v {/eq} are evaluated by applying Chain rule and treating another variable a constant, where

{eq}\dfrac{\partial f}{\partial u} = \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial u} + \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial u} \\ \dfrac{\partial f}{\partial v} = \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial v} + \dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial v} \\ {/eq}

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Given {eq}z = e^{x^2y} {/eq} where {eq}x = \sqrt{uv} {/eq} and {eq}y=\frac1v {/eq}.

Then, we evaluate the partial derivatives of {eq}z {/eq} with...

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Learn more about this topic:

The Chain Rule for Partial Derivatives


Chapter 14 / Lesson 4

This lesson defines the chain rule. It goes on to explore the chain rule with partial derivatives and integrals of partial derivatives.

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