Given the function: f(x, y) = x sin y + xy cos x, compute partial f / partial x , partial f /...

Question:

Given the function: {eq}\displaystyle f(x,\ y) = x \sin y + xy \cos x {/eq}, compute {eq}\displaystyle \dfrac {\partial f }{\partial x},\ \dfrac {\partial f} {\partial y},\ \dfrac {\partial^2 f }{\partial x^2},\ \dfrac {\partial^2 f }{\partial y^2},\ \dfrac {\partial^2 f }{\partial x \partial y} {/eq}, and {eq}\displaystyle \dfrac {\partial^2 f }{\partial y \partial x} {/eq}.

Partial Derivatives:

Let us consider a function {eq}\displaystyle f\left(x, y\right) {/eq} of two variable {eq}\displaystyle x~\mbox{and}~y {/eq}. Then first order partial derivatives of {eq}\displaystyle f {/eq} with respect to {eq}\displaystyle x {/eq} is {eq}\displaystyle \dfrac {\partial f }{\partial x} {/eq} and with respect to {eq}\displaystyle y {/eq} is {eq}\displaystyle \dfrac {\partial f }{\partial y} {/eq}.

And second order partial derivatives are {eq}\displaystyle \dfrac {\partial^2 f }{\partial x^2}=\dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x}\right),\ \dfrac {\partial^2 f }{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial y}\right),\ \dfrac {\partial^2 f }{\partial x \partial y}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right), \ \dfrac{\partial^2 f}{\partial y \partial x}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right) {/eq}, where {eq}\displaystyle \dfrac {\partial^2 f }{\partial x \partial y}, \ \dfrac{\partial^2 f}{\partial y \partial x} {/eq} are called mixed partial derivatives.

Answer and Explanation: 1

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Given, {eq}\displaystyle f(x,\ y) = x \sin y + xy \cos x {/eq}.

Then, {eq}\displaystyle \dfrac {\partial f }{\partial x}=\sin y+y \cos x-xy\sin...

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Partial Derivative: Definition, Rules & Examples

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Chapter 18 / Lesson 12
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What is a Partial Derivative? Learn to define first and second order partial derivatives. Learn the rules and formula for partial derivatives. See examples.


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