- #1
DiracPool
- 1,243
- 516
I'm trying to understand how the derivative of this function:
x=ρcosθ
Becomes this:
dx=−ρsinθdθ+cosθdρ
First off I'm guessing that x is a function of both ρ AND cosθ, or else we wouldn't be using the product rule in the first place..Am I correct? So how could we write this in functional notation? Like so?
x(ρ,cosθ)= ρcosθ
Then might the next step of taking the derivative of x with respect to ρ and cosθ look like this?
dx/(dρ dθ) = −ρsinθ+cosθ
Moving the denominator on the left side to the right could yield the answer:
dx=−ρsinθdθ+cosθdρ
But how do we know to put the dθ next to the sin term and the dρ next to the cos term? Perhaps I'm working this out wrong?
To put it another way, if we took the derivative of the function x=ρcosθ with respect to each of the variables separately, we'd get
dx/dρ = cosθ and dx/dθ = -ρsin(θ)
But I guess I'm missing how we combine these separate operations to get the equation:
dx=−ρsinθdθ+cosθdρ
Perhaps we are not even using the product rule here?
x=ρcosθ
Becomes this:
dx=−ρsinθdθ+cosθdρ
First off I'm guessing that x is a function of both ρ AND cosθ, or else we wouldn't be using the product rule in the first place..Am I correct? So how could we write this in functional notation? Like so?
x(ρ,cosθ)= ρcosθ
Then might the next step of taking the derivative of x with respect to ρ and cosθ look like this?
dx/(dρ dθ) = −ρsinθ+cosθ
Moving the denominator on the left side to the right could yield the answer:
dx=−ρsinθdθ+cosθdρ
But how do we know to put the dθ next to the sin term and the dρ next to the cos term? Perhaps I'm working this out wrong?
To put it another way, if we took the derivative of the function x=ρcosθ with respect to each of the variables separately, we'd get
dx/dρ = cosθ and dx/dθ = -ρsin(θ)
But I guess I'm missing how we combine these separate operations to get the equation:
dx=−ρsinθdθ+cosθdρ
Perhaps we are not even using the product rule here?