U1S4V09 Derivative of a sum - proof.txt

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We said that if h of x equals f of x plus g of x for all x,
then h prime of x is going to be f prime of x plus g prime of x.
In other words, the derivative of the sum
is the sum of derivatives.
So let's prove this.
We'll start just with the definition for h prime of x.
And we'll use the delta x definition,
so that's the limit as delta x goes
to 0 of h of x plus delta x minus h of x all over delta x.
And that's going to be the limit as delta x goes to 0.
And now we just use what we know about h.
So h of anything, that's just going
to be the sum of the corresponding f
and the corresponding g.
So when we see h of x plus delta x,
that's just f of x plus delta x plus g of x plus delta x.
And then we're subtracting h of x, which is f of x plus g of x.
And all of this is over delta x.
So this is a little bit complicated,
but let's just push all the f's to one side
and then put all the g's on the other side.
So we're going to have f of x plus delta x
minus f of x plus g of x plus delta x minus g of x.
And both of these are over delta x.
I'm just going to make it two separate fractions here.
And we're taking the limit as delta x goes to 0.
So we have the limit of this sum.
And now if you look at the two parts
as we're taking the limit as delta x goes to 0,
this first part is just f prime of x.
That's the definition.
And this second part, similarly, is g prime of x.
So this limit of the sum ends up just being f prime of x plus g
prime of x.
And it works.
Hurray.