A really fast wuestion about a partial derivative

In summary, the homework statement is unclear about what the derivative is supposed to be. It is possible that the derivative is supposed to be (alpha)2x, but this is not clear. It is also possible that the derivative is supposed to be (alpha)2y, but this is also not clear. If the units matter during differentiation, then the derivative will involve (N/m2)/m, or N/m3.
  • #1
vande060
186
0

Homework Statement



find the partial derivative with respect to x

Homework Equations



f(x) = (alpha)x^2

alpha is a just to represent units N/m^2

The Attempt at a Solution



well, i know that during a partial derivative all variable but the one of interest, x in this case, are held constant. so i don't understand why the derivative is not (alpha)2x. it is 0 according to my prof, and i don't understand why.

if it helps this is a conservative test in physics

f = (x^2i + y^2j)alpha

i want to make sure that the derivative of these two components are equal, they obviously cannot be if one is alpha2x and the other is alpha2y
 
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  • #2
do the units matter during differentiation

if i carry out the chain rule 1/m^2 becomes 1/m^3 * m' , and the derivative of m would be zero because it is constant. this would then make the entire term 0
 
  • #3
Wait.. Is your prof saying that d/dx alpha*x^2 = 0?
 
  • #4
vande060 said:
do the units matter during differentiation

if i carry out the chain rule 1/m^2 becomes 1/m^3 * m' , and the derivative of m would be zero because it is constant. this would then make the entire term 0
You don't differentiate units. If the units of the original function are N/m2, and you differentiate with respect to x, the difference quotient will involve (N/m2)/m, or N/m3.
 
  • #5
Inferior89 said:
Wait.. Is your prof saying that d/dx alpha*x^2 = 0?
Yeah, that doesn't make any sense to me, either.
 
  • #6
Mark44 said:
You don't differentiate units. If the units of the original function are N/m2, and you differentiate with respect to x, the difference quotient will involve (N/m2)/m, or N/m3.

Assuming that x is measured in meters.
 
  • #7
i guess i will have to ask him how he got that answer, i still have no idea
 

Related to A really fast wuestion about a partial derivative

1. What is a partial derivative?

A partial derivative is a type of mathematical derivative that measures the rate of change of one variable with respect to another in a multivariable function.

2. How is a partial derivative different from a regular derivative?

A partial derivative focuses on the rate of change of one variable while holding all other variables constant, whereas a regular derivative considers the overall rate of change of a function.

3. When is a partial derivative useful?

A partial derivative is useful in fields such as physics, engineering, and economics where a system may have multiple variables that affect each other.

4. How is a partial derivative calculated?

To calculate a partial derivative, you take the derivative of the function with respect to the variable of interest and treat all other variables as constants. This is done using the traditional rules of differentiation.

5. What are some real-world applications of partial derivatives?

Some real-world applications of partial derivatives include optimization problems in economics, determining marginal cost and revenue in business, and analyzing the rate of change in physical systems such as heat transfer or fluid dynamics.

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