# Compute Higher Order Mixed Derivative.

• smerhej
In summary, we are given a function f(u,v) and g(x,y) = (x^2 + y^4, xy). Using the chain rule and product rule, we can find the derivative of d(f o g)/dxdy. After differentiating with respect to y and x, we can find the second order derivatives and plug in the given values for f_{v} (5,2) = 1, f_{uu} (5,2) = 2, f_{vv} (5,2) = -2 and f_{uv} (5,2) = 1 to solve for d^2(f o g)/dxdy at (2,1).
smerhej

## Homework Statement

Let f(u,v) be an infinitely differentiable function of two variables, and let g(x,y) = (x^2 + y^4, xy). If f$_{v}$ (5,2) = 1, f$_{uu}$ (5,2) = 2, f$_{vv}$ (5,2) = -2 and f$_{uv}$ (5,2) = 1, computer d^2(f o g)/dxdy at (2,1)

The Chain Rule

## The Attempt at a Solution

I set u = x^2 + y^4 and v = xy . Differentiating df/dx gives 2x(df/du) + y(df/dv) .[ df/dy gives 4y^3(df/du) + x(df/dv) but I don't think that's relevant ]. My idea from there was to use the results stated in the question to fill in these unknown derivatives.

Thanks!

Yes, $\partial f/\partial x= 2x \partial f/\partial u+ y\partial f/\partial v$.

Now, differentiate that with respect to y:
$2x\partial^2 f/\partial u^2(4y^3)+ ...$

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Use (derive?) Faà di Bruno's formula.
Just include all possibilities
or do
chain rule
product rule (twice)
chain rule (twice)

$$(f \circ g)_{xy}=(f_u u_x+f_v v_x)_y \\ = f_u u_{xy}+f_v v_{xy}+(f_u)_y u_x+(f_v)_y v_x\\ =f_u u_{xy}+f_v v_{x y}+f_{uu} u_x u_y+f_{uv} u_x v_y+f_{vu} v_x u_y+f_{vv} v_x v_y$$
where subscripts denote differentiation and g(x,y)=(u,v)

Ok so I'm pretty sure I got the derivative worked out. First we're trying to figure out d(f o g)/dxdy. So differentiate with respect to y first then x. Doing so gives:

[ df/dy = 4y3 df/du + xdf/dv ]

Differentiating with respect to x gives:

[ 4y3d/dx(df/du) + df/dv +d/dx(df/dv) ]

Then, differentiating d/dx(df/du) and d/dx(df/dv), and putting those back into the equation above gives:

4y3d2f/du22x + d2f/dudvy +df/dv + d2f/dvdu2x + d2f/dv2y

If anyone wants to link me a "how to write math properly on this forum guide" that would be swell..

## 1. What is a higher order mixed derivative?

A higher order mixed derivative is a mathematical concept used in calculus to describe the rate of change of a function with respect to multiple variables. It involves taking the derivative of a function with respect to one variable, and then taking the derivative of that result with respect to another variable.

## 2. How is a higher order mixed derivative calculated?

A higher order mixed derivative is calculated by taking the derivative of a function with respect to one variable, and then taking the derivative of that result with respect to another variable. This process can be repeated for multiple variables to obtain higher order mixed derivatives.

## 3. What is the difference between a first order and higher order mixed derivative?

The main difference between a first order and higher order mixed derivative is the number of variables involved. A first order mixed derivative involves taking the derivative of a function with respect to two variables, while a higher order mixed derivative involves taking the derivative with respect to three or more variables.

## 4. How is a higher order mixed derivative used in real world applications?

Higher order mixed derivatives are used in various real world applications, such as physics, engineering, and economics. They can be used to describe the rate of change of a system with multiple variables and can help predict future behavior of the system.

## 5. Are there any limitations to using higher order mixed derivatives?

One limitation of using higher order mixed derivatives is that they can become increasingly complex and difficult to calculate as the number of variables increases. Additionally, they may not always accurately describe the behavior of a system in the real world and must be carefully interpreted and applied in specific situations.

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