# Help understanding the Chain Rule book for derivatives

1. Dec 24, 2014

### MidgetDwarf

After completing calculus 2 with an A I now realize I know nothing of mathematics. We used stewart calculus and I did not really like it, due to alot of hand waiving.

I got an older edition of thomas calculus with analytic geometry 3rd ed, and so far I'm having a blast learning proofs from this book and the insight the author has.

I'm confused on p 78 from thomas about the proof of the chain rule for derivatives.

We are given x=f (t) and y=g (t) which are parametric equations.

If y=F (x) is a differential able function of x and x=f (t) is a differentiable function of t, then y=F [f(t)]=g (t)
is a differentiable function of t and g'(t)=F'(x) f'(t), or in other words dy/dx=(dy/dx)(dx/dt).

My attempt,

G: y=F [f (t) ]

Take the derivative of outside y=F (x) with respect to x.

Here is what is confusing me. If I take the derivative of the above function. It is dy/dx=F'(x)

Then taking the derivative of the inside function x=f (t) with respect to t.

Becomes dx/dt=f'(t)

Now we cancel out terms:(dy/dx)(dx/dt)=(dy/dt) ?

Can anyone she'd light on this please.

There is more to the proof where we have to use Linearization proof ( this part I understand).
Sorry for the winded post I'm pulling my beard out the night before Christmas trying to figure it out.

2. Dec 24, 2014

### Fredrik

Staff Emeritus
There's no cancellation in $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$. dx, dy, dt are usually left undefined. I don't know a way to define them that enables you to view the chain rule as an example of a cancellation.

I prefer the following presentation of the chain rule:

I will never call f(x) a function. f is the function. f(x) is its value at x. So f(x) is typically a number.

Let f and g be functions such that the range of g is a subset of the domain of f. Then we can define a new function $f\circ g$ by $(f\circ g)(x)=f(g(x))$ for all x in the domain of g. The function $f\circ g$ is called the composition of f and g. The chain rule says that
$$(f\circ g)'(x)=f'(g(x))g'(x)$$ for all x in the domain of g.

There are many ways to prove this result. Unfortunately they're all kind of difficult. The straightforward proof (a direct application of the definitions of limit and derivative) is kind of complicated, and the short proofs are tricky. See posts #6 and #7 here for a non-rigorous argument followed by a proof.

3. Dec 25, 2014

### MidgetDwarf

Thank you Mr Fredrick, for the alternative ways to prove the chain rule. It seems that It is currently beyond my level to fully grasp the proof. I will study it a bit more to see if I can get more insight. I really appreciate it.