Partial derivative interpretation

[Celso]

How do I interpret geometrically the partial derivative in respect to a constant of a function such as $\frac{ \partial}{\partial c} (acos(x) + be^x + c)^2$?

 

[Stephen Tashi]

There are some contexts where people call a letter in an expression a "constant" and yet intend the letter to function as a variable or consider it a variable for part of their discussion and then treat it as an actual constant in other parts. So you should explain the complete context for the formula you are asking about.

If "c" is actually a constant, like c = 4.73 then the partial derivative of a function with respect to "c" is zero. Does that need a geometric interpretation?

If we are momentarily pretending "c" is a variable, then the geometric interpretation is like the geometric interpretation for other variables.

 

[BvU]

You look at the variation in the function value while keeping everything constant, except $c$

To work out your example you'll still need the chain rule: to take the derivative of $(D + c)^2\$ (where $D$ is a constant) wrt $c$, you'd get $$2(D+c)\; {\partial \over \partial c } (D+c) = 2(D+c) \ $$ or $\ 2 (\arccos x + be^x + c)$

Geometrically you'd be working in the plane where $a, b\$ and $x$ are constant and the horizontal axis is the $c$ axis, the vertical axis is the function value. The derivative is the tangent of the slope.

To avoid miscommunication like what Stephen hints at, you could use the term 'parameter of the function' instead of 'constant of the function'. But I grant you that it's just a convention.

 

[jedishrfu]

I guess you're trying to see how the function varies as you change the constant.

As an example $y=x^2 + c$ as c varies the function's shape stays the same but it slides up and down on the graph of x vs y.

Next consider $y=(x+c)^2$ now as c varies the function again stays that same but slides back and forth on the graph of x vs y.

I know that sometimes to integrate a given function its easier to differentiate the constant of a simpler function of the same form that you already know the integral to in order to discover the integral of the given function.

From Prof Nearing's Math Methods book:

http://www.physics.miami.edu/~nearing/mathmethods/mathematical_methods-three.pdf

1.2 Parametric Differentiation

The integration techniques that appear in introductory calculus courses include a variety of methods of varying usefulness. There’s one however that is for some reason not commonly done in calculus courses: parametric differentiation. It’s best introduced by an example:

...see section 1.2 of the book...

You could integrate by parts n times and that will work. Instead of this method, do something completely different. Consider the integral of $xe^{\alpha x}$ It has the parameter $\alpha$ in it. Differentiate with respect to $\alpha$.

...

The idea of this method is to change the original problem into another by introducing a parameter. Then differentiate with respect to that parameter in order to recover the problem that you really want to solve. With a little practice you’ll find this easier than partial integration.

Also see problem 1.47 for a variation on this theme. Notice that I did this using definite integrals. If you try to use it for an integral without limits you can sometimes get into trouble. See for example problem 1.42

(I had to paraphrase the books discussion due to poor latex skills on my part)