Derivative defined as a limit
- Thread starter courteous
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Use the quotient rule:
Let's assume f is differentiable at x and \(\displaystyle f(x)\neq 0\), then \(\displaystyle \frac{1}{f(x)}\) is differentiable at x and
\(\displaystyle \frac{d}{dx}\left[\frac{1}{f(x)}\right]=\frac{f(x)(0)-(1)f'(x)}{[f(x)]^{2}}=\frac{-f'(x)}{[f(x)]^{2}}=\frac{d}{dx}\left[\frac{1}{f(x)}\right]\)
Let's assume f is differentiable at x and \(\displaystyle f(x)\neq 0\), then \(\displaystyle \frac{1}{f(x)}\) is differentiable at x and
\(\displaystyle \frac{d}{dx}\left[\frac{1}{f(x)}\right]=\frac{f(x)(0)-(1)f'(x)}{[f(x)]^{2}}=\frac{-f'(x)}{[f(x)]^{2}}=\frac{d}{dx}\left[\frac{1}{f(x)}\right]\)
What if you pretend you do not know quotient rule?
0)Preeminently, derivative is defined as a limit of a differential quotient: \(\displaystyle \lim_{h\to 0}{f(a+h)-f(a)\over h}\).
1)Then, textbook proves (using above definition all the way): \(\displaystyle [(f+g)(x)]'=f'(x)+g'(x)\).
2)Then, it proves the product rule.
3)Then, it builds on previous two ("sum rule" and product rule; still using the definition) proving quotient rule: \(\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]=\lim_{h\to0}\frac{1}{h}[\frac{f(x)-f(x+h)}{f(x+h)f(x)}]=...=-\frac{f'(x)}{f^2(x)}\)
:!: Oblivious of quotient rule, how would you show that \(\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]=...\) :?:
0)Preeminently, derivative is defined as a limit of a differential quotient: \(\displaystyle \lim_{h\to 0}{f(a+h)-f(a)\over h}\).
1)Then, textbook proves (using above definition all the way): \(\displaystyle [(f+g)(x)]'=f'(x)+g'(x)\).
2)Then, it proves the product rule.
3)Then, it builds on previous two ("sum rule" and product rule; still using the definition) proving quotient rule: \(\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]=\lim_{h\to0}\frac{1}{h}[\frac{f(x)-f(x+h)}{f(x+h)f(x)}]=...=-\frac{f'(x)}{f^2(x)}\)
:!: Oblivious of quotient rule, how would you show that \(\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]=...\) :?:
D
Deleted member 4993
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Or you can think in the following way:
Let
\(\displaystyle \frac{1}{f(x)} \, = \, G(x)\)
\(\displaystyle \frac{1}{f(x+h)} \, = \, G(x+h)\)
\(\displaystyle [\frac{1}{f(x)}]' \, = \, G'(x) \, = \, \lim_{h\to 0}\frac{G(x+h) \, - \, G(x)}{h} \, = \, \lim_{h\to 0}\frac{\frac{1}{f(x+h)} \, - \, \frac{1}{f(x)}}{h}\)
No quotient rule needed...
Let
\(\displaystyle \frac{1}{f(x)} \, = \, G(x)\)
\(\displaystyle \frac{1}{f(x+h)} \, = \, G(x+h)\)
\(\displaystyle [\frac{1}{f(x)}]' \, = \, G'(x) \, = \, \lim_{h\to 0}\frac{G(x+h) \, - \, G(x)}{h} \, = \, \lim_{h\to 0}\frac{\frac{1}{f(x+h)} \, - \, \frac{1}{f(x)}}{h}\)
No quotient rule needed...
