Copyright

Suppose u(x,y), v(x,y) are functions that satisfy u(0,1)= 2 \frac{\partial u}{\partial x}(0,1)=...

Question:

Suppose u(x,y), v(x,y) are functions that satisfy u(0,1)= 2 {eq}\frac{\partial u}{\partial x}(0,1)= 0.5919,\ \frac{\partial u}{\partial y}(0,1)= 1.3961 {/eq}, v(0,1)= 1 {eq}\frac{\partial v}{\partial x}(0,1)= 0.6850,\ \frac{\partial v}{\partial y}(0,1)= 0.6751 {/eq}, {eq}\frac{\partial^2 u}{\partial x\partial y}(0,1)= 1.7286,\ \frac{\partial^2 v}{\partial x\partial y}(0,1)= 1.4093 {/eq}.

For the function {eq}f= \sin(u^2) - \cos(2uv - v^2) {/eq}, find the value of {eq}\frac{\partial^2 f}{\partial x\partial x}at\ x= 0\ and\ y= 1 {/eq}.

Chain Rule:

For multivariable functions, we can apply a method of organization to the chain rule in order to simplify computation. Beginning at the parent function and ending at the variable in question, we consider all equation paths we can take to obtain the given variable. We then perform the relevant partial derivative along each path and multiply each piece. Finally, we add together the results from each path.

Answer and Explanation: 1

Become a Study.com member to unlock this answer! Create your account

View this answer

We first find the derivatives symbolically. We have

{eq}\displaystyle \begin{align*} \frac{df}{dx} &= \frac{df}{du}\frac{du}{dx} +...

See full answer below.


Learn more about this topic:

Loading...
The Chain Rule for Partial Derivatives

from

Chapter 14 / Lesson 4
35K

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


Related to this Question

Explore our homework questions and answers library