I am not too grounded in differentiation but today, I was posed with a supposedly easy question w=f(x,y)=x2+y2 where x=rsinθ and y=rcosθ requiring the solution to ∂w/∂r and ∂w/∂θ. I simply solved the former using the trig identity sin2θ+cos2θ=1, resulting to ∂w/∂r=2r.
However I was told that this solution could not be applied to this question because I should be solving for the total derivative. I could not find any good resource online to explain clearly to me the difference between a normal derivative and a total derivative and why my solution here was wrong. Is there anyone who could explain the difference to me using a practical example? Thanks!
Answer
The key difference is that when you take a partial derivative, you operate under a sort of assumption that you hold one variable fixed while the other changes. When computing a total derivative, you allow changes in one variable to affect the other.
So, for instance, if you have f(x,y)=2x+3y, then when you compute the partial derivative ∂f∂x, you temporarily assume y constant and treat it as such, yielding ∂f∂x=2+∂(3y)∂x=2+0=2.
However, if x=x(r,θ) and y=y(r,θ), then the assumption that y stays constant when x changes is no longer valid. Since x=x(r,θ), then if x changes, this implies that at least one of r or θ change. And if r or θ change, then y changes. And if y changes, then obviously it has some sort of effect on the derivative and we can no longer assume it to be equal to zero.
In your example, you are given f(x,y)=x2+y2, but what you really have is the following:
f(x,y)=f(x(r,θ),y(r,θ)).
So if you compute ∂f∂x, you cannot assume that the change in x computed in this derivative has no effect on a change in y.
What you need to compute instead is dfdθ and dfdr, the first of which can be computed as:
dfdθ=∂f∂θ+∂f∂xdxdθ+∂f∂ydydθ
Attribution
Source : Link , Question Author : Chibueze Opata , Answer Author : Emily