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

. I could not find any good resource online to explain clearly to me the difference between atotal derivativenormalderivative and atotalderivative and why my solution here waswrong. 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However, if x=x(r,θ) and y=y(r,θ), then the assumption that y stays constant when x changes is no longer valid.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*