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dcfcviper · Dec 10, 2007, 09:22 AM

Differentiate $\frac{d}{dx}[\frac{1}{g(x)}]$ from first principles and use the result to derive the product rule assuming the product rule to be true.

I've differentiated it using the quotient rule (get $\frac{-g'(x)}{(g(x))^2}$ ) to use as a check and also by the chain rule but cannot reach the answer through first principles or derive the quotient rule using the answer I got for the first part by a different method.

Won't post all the workings, but I started with the definition of differentiation from first principles and let ${f(x)=\frac{1}{g(x)}$ and worked through it but the closest I think I get is $\frac{0}{(g(x))^2}$ and I think I even made errors with that.

So any help would be appreciated :)

Capuchin · Dec 10, 2007, 10:05 AM

Use the fact that $\frac{1}{x} = x^{-1}$ . That should make it a form that is easy to differentiate.

ebaines · Dec 10, 2007, 10:33 AM

Start with the definition of the derivative, like this:
$<br /> \frac {d(1/g(x))} {dx} =\displaystyle\lim_{h \to 0} $ \frac { \frac 1 {(g(x+h)} - \frac 1 {g(x)}} h $$

Massage this a bit and you get:

$<br /> - \displaystyle\lim_{h \to 0} $ \frac {g(x+h) - g(x)} h $ \cdot \frac 1 {g(x+h)g(x)}<br />$

Notice the first term is the definition of g'(x). Can you take it from here?

dcfcviper · Dec 10, 2007, 10:49 AM

Capuchin - Tried and failed that way somehow

ebaines - Thanks, helps a bit.

dcfcviper · Dec 10, 2007, 02:10 PM

Now managed to complete the whole question including deriving the quotient rule, many thanks again guys :thup: