What exactly is the difference between a derivative and a total derivative?

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 fx, you temporarily assume y constant and treat it as such, yielding fx=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However, if x=x(r,θ) and y=y(r,θ), then the assumption that y stays constant when x changes is no longer valid.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 fx, 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=fθ+fxdxdθ+fydydθ

Attribution
Source : Link , Question Author : Chibueze Opata , Answer Author : Emily

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