1. Differentials

edit - solved...haha i just realised you differentiate with respect to one first, then add the differential with respect to the other.

Hi everyone,

I just need some help on a question that I'm currently trying to solve. I'm doing Summer Maths at uni at the moment but I haven't quite grasped this concept.

The question is telling me to calculate a differential, given an equation.

A = xy

Find dA.

I know for a fact that the answer is dA = ydx + xdy, but I am unsure as to how the lecturer got to that process because he stated it was 'fairly self explanatory'. I guess I didn't have the guts to ask infront of a hundred people so here I am...haha

I don't expect a complete answer...moreso some guidance or references that will help me along the way.

2. Re: [SOLVED]Differentials

Originally Posted by ivanyo
edit - solved...haha i just realised you differentiate with respect to one first, then add the differential with respect to the other.

Hi everyone,

I just need some help on a question that I'm currently trying to solve. I'm doing Summer Maths at uni at the moment but I haven't quite grasped this concept.

The question is telling me to calculate a differential, given an equation.

A = xy

Find dA.

I know for a fact that the answer is dA = ydx + xdy, but I am unsure as to how the lecturer got to that process because he stated it was 'fairly self explanatory'. I guess I didn't have the guts to ask infront of a hundred people so here I am...haha

I don't expect a complete answer...moreso some guidance or references that will help me along the way.

Differentiate both sides with respect to x. Since y is a function of x, on the RHS you need to use the product rule...

3. Re: Differentials

Suppose x and y are functions of some paramter, t: x= x(t), y= y(t). Then we can think of f(x,y) as a function of the single variable t.

By the chain rule, $\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}$.

The differential, for a function of a single variable, is defined by $df= \frac{df}{dt}dt$ so we have here
$df= \frac{df}{dx}dt= \left(\frac{\partial f}{\partial x}\frac{dx}{dt}+ \frac{\partial f}{\partial y}\frac{dy}{dt}\right)dt$
$= \frac{\partial f}{\partial x}\left(\frac{dx}{dt}dt\right)+ \frac{\partial f}{\partial y}\left(\frac{dy}{dt}dt\right)= \frac{\partial f}{\partial x}{dx}+ \frac{\partial f}{\partial y}dy$
which is now independent of t.