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**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 a lot 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) ]

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.