Question

Prove the quotient rule using chain-rule and the derivative of 1/x

Answer 1

You aren't allowed the product rule I assume

Answer 2

@mhchen

Answer 3

Then it's trivial

Answer 4

d/dx u(x)/v(x)=d/dx u(x) * (1/v(x)) = u' *1/v + u*(-1/v^2) and you can do the last algebraic step

Answer 5

And stop thinking in terms of limits that's like trying to do multiplication with repeated addition

Answer 6

Unless you have to use limits @mhchen

Answer 7

In that case expand each derivative into a difference quotient and show the expression holds

Answer 8

Otherwise you can show the product and chain rule hold as well as the derivative of 1/u

Answer 9

Then, redefine the derivative notation as the difference quotient, and just finish what I got you to.

Answer 10

Derivative of 1/x is this:
\[\lim_{x \rightarrow c}\frac{\frac{1}{x}-\frac{1}{c}}{x-c} = \lim_{x \rightarrow c}\frac{c-x}{xc(x-c)}\]

Now then, derivative of
\[(\frac{f(c)}{g(c)})' = f'(c)\frac{1}{g(c)}+f(c)(\frac{1}{g(c)})'\]

\[\lim_{x \rightarrow c}\frac{f(x)-f(c)}{(x-c)(g(c)} + f(c)\lim_{x \rightarrow c}\frac{c-g(c)}{g(c)c(g(c)-c)}*\lim_{x \rightarrow c}\frac{g(x)-g(c)}{x-c}\]
I used chain rule here


\[\lim_{x \rightarrow c} (\frac{(f(x)-f(c))g(c)}{(x-c)g(c)^{2}} - f(c)\frac{g(x)-g(c)}{g(c)c(x-c)})\]

\[\frac{\lim_{x \rightarrow c}(\frac{f(x)-f(c)}{x-c})g(c)}{g(c)^{2}}-\frac{f(c)\lim_{x \rightarrow c}(\frac{g(x)-g(c)}{x-c})}{g(c)c}\]

My entire work using limits, product rule, and chain rule is here

Answer 11

I can't follow that. Just rewrite each step of my work as a limit and you can show that it holds. Simplify intermediate steps as I did so you don't get limit spaghetti

Answer 12

Also your class Prof is a goofball they might as well make you derive the formula via the epsilon Delta limit definition -.-

Answer 13

I never dared to try to understand the epsilon delta limit definition or the 'official definition'

Answer 14

Do you need to use limits or can you compose something through if you know what I mean?

Answer 15

The epsilon Delta def is only one way to define it. It's popular in real analysis but not really necessary and impedes understanding when you think about a lot of limits... It's like trying to do counting with the peano axioms

Answer 16

Also @mhchen I'd write out the Sol for you but I'm on mobile and typing is already hard enough