A really fast wuestion about a partial derivative

[vande060]

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

 

[vande060]

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

 

[Inferior89]

Wait.. Is your prof saying that d/dx alpha*x^2 = 0?

 

[Mark44]

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.

 

[Mark44]

Wait.. Is your prof saying that d/dx alpha*x^2 = 0?

Yeah, that doesn't make any sense to me, either.

 

[Inferior89]

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.

 

[vande060]

i guess i will have to ask him how he got that answer, i still have no idea