Suppose that f(x, y)=e^{x y}, x(u, v)=3 u \sin v and y(u, v)=4 v^{2} u. For g(u, v)=f(x(u, v),...


Suppose that {eq}null{/eq} and {eq}null{/eq}. For {eq}null{/eq} Find the partial derivates. a) {eq}null{/eq} b){eq}null{/eq}

Chain Rule of Partial Derivatives:

The chain rule of partial derivatives is used to find the partial derivative of a two-variable function {eq}f(x,y) {/eq} where each of the two variables is defined as a function of two variables {eq}u {/eq} and {eq}v {/eq}. This rule states:

$$\dfrac{\partial \mathbf{f}}{\partial \mathbf{u}}=\dfrac{\partial \mathbf{f}}{\partial \mathbf{x}} \cdot \dfrac{\partial \mathbf{x}}{\partial \mathbf{u}}+\dfrac{\partial \mathbf{f}}{\partial \mathbf{y}} \cdot \dfrac{\partial \mathbf{y}}{\partial \mathbf{u}} \\ \dfrac{\partial \mathbf{f}}{\partial \mathbf{v}}=\dfrac{\partial \mathbf{f}}{\partial \mathbf{x}} \cdot \dfrac{\partial \mathbf{x}}{\partial \mathbf{v}}+\dfrac{\partial \mathbf{f}}{\partial \mathbf{y}} \cdot \dfrac{\partial \mathbf{y}}{\partial \mathbf{v}} $$

Answer and Explanation:

The given function is:

$$f(x, y)=e^{x y} $$

We will find its partial derivatives with respect to {eq}x {/eq} and {eq}y {/eq}.

$$\dfrac{\partial \mathbf{f}}{\partial \mathbf{x}}=ye^{x y}\\ \dfrac{\partial \mathbf{f}}{\partial \mathbf{y}}=xe^{x y} $$

The given two functions are:

$$x(u, v)=3 u \sin v \\ y(u, v)=4 v^{2} u $$

We will find the partial derivatives of each of these functions with respect to {eq}u {/eq} and {eq}v {/eq}.

$$\dfrac{\partial \mathbf{x}}{\partial \mathbf{u}} = 3 \sin v \\ \dfrac{\partial \mathbf{x}}{\partial \mathbf{v}} = 3u \cos v \\ \dfrac{\partial \mathbf{y}}{\partial \mathbf{u}} = 4v^2(1)=4v^2 \\ \dfrac{\partial \mathbf{y}}{\partial \mathbf{v}} = 4u(2v)=8uv $$

Now by the chain rule of partial derivatives:

$$\begin{align} (a) \dfrac{\partial \mathbf{f}}{\partial \mathbf{u}}&=\dfrac{\partial \mathbf{f}}{\partial \mathbf{x}} \cdot \dfrac{\partial \mathbf{x}}{\partial \mathbf{u}}+\dfrac{\partial \mathbf{f}}{\partial \mathbf{y}} \cdot \dfrac{\partial \mathbf{y}}{\partial \mathbf{u}} \\ &= ye^{xy} (3 \sin v) + x e^{xy} (4v^2) \\ & =\boxed{\mathbf{ 3ye^{xy} \sin v + 4xv^2 e^{xy}}} \end{align} $$

$$\begin{align} (b) \dfrac{\partial \mathbf{f}}{\partial \mathbf{v}}&=\dfrac{\partial \mathbf{f}}{\partial \mathbf{x}} \cdot \dfrac{\partial \mathbf{x}}{\partial \mathbf{v}}+\dfrac{\partial \mathbf{z}}{\partial \mathbf{y}} \cdot \dfrac{\partial \mathbf{y}}{\partial \mathbf{v}}\\ & = ye^{xy} (3u \cos v) + xe^{xy} (8uv) \\ & =\boxed{\mathbf{ 3uy e^{xy} \cos v + 8xuve^{xy} }} \end{align} $$

Learn more about this topic:

The Chain Rule for Partial Derivatives

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 4

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