U2S4V11 Chain Rule vs. Quotient Rule.txt

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So here's our function, and we want to take its derivative.
Now, you could use the quotient rule, of course.
There's nothing wrong with that, but there
are some people who can't remember or don't want
to remember the quotient rule.
And the good news for such people
is that they don't have to.
You can always differentiate a quotient using the chain rule
instead.
So the way that that works is by thinking
of this using the reciprocal function, which I'll call r.
So r of t is just 1 over t or t to the minus 1.
And then our h becomes not r of x, but r of u,
where u is this expression, x squared plus x.
Now, to take the derivative, we need the chain rule.
So we take the derivative of the outer function r applied
to u times the derivative of u.
And the derivative of r is just using the power rule minus 1
over t squared.
And so when we apply that to u that's minus 1 over u squared.
And then we need the derivative of u, so that's 2x plus 1.
And putting everything in terms of x, we get minus 1
over quantity x squared plus x squared times 2x plus 1.
So that's it.
No quotient rule needed.
Now, if you want to avoid the quotient rule in general,
you're going to need not just the chain rule,
but also the product rule.
So let's erase this, and let's suppose that h of x
is the quotient f of x over g of x.
But if we don't want to use the quotient rule,
then we should rewrite this as f of x
times the reciprocal of g of x.
And now we can see that we can start differentiating
this using the product rule.
So we want the first function, so that's
f of x times the derivative of the second function.
So that's r of g of x prime plus the second function
times the derivative of the first function.
So we get f of x times, and then for this derivative
we need the chain rule here, so we
need the derivative of r applied to the inner function g of x.
So that's minus 1 over g of x squared times the derivative
of the inner function.

And then over here, I'm going to rewrite r of g of x as 1
over g of x times f prime of x.
And so now if we put everything over the common denominator,
which is g of x squared then on top we're going to get minus f
of x times g prime of x.
And then the term on the right is now
f prime of x times g of x.
And this is exactly the quotient rule.
So if you ever forget it, now you
know how to work without it.
Sound good?