This should be fairly simple but I'm missing something. The way that I derived this equation was $$f(x)=\frac{3x-6}{x}$$ $$f'(x)=\frac{3(1x^0)-0}{1x^0}=\frac{3}{1}=3$$
What am I doing wrong?
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https://en.wikipedia.org/wiki/Quotient_rule – Hans Lundmark Aug 25 '18 at 08:58
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4$$f(x) = x$$ is a function! – Davide Morgante Aug 25 '18 at 09:06
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You cannot differentiate numerator and denominator separately.
While you could use the quotient rule $(*)$, I usually try to avoid it when that's easily done: $$\left(\frac{3x-6}{x}\right)'=\left(3-\frac{6}{x}\right)'=\left(3-6x^{-1}\right)'=6x^{-2}=\frac{6}{x^2}$$
With the quotient rule $(*)$: $$\left(\frac{3x-6}{x}\right)'=\frac{(3x-6)'x-(3x-6)x'}{x^2}=\frac{3x-(3x-6)}{x^2}=\frac{6}{x^2}$$
$(*)$ the quotient rule tells you how to differentiate fractions: $$\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$ Note how this is (very) different from $\frac{f'(x)}{g'(x)}$...!
StackTD
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By quotient rule
$$f’(x)=\frac{3x-(3x-6)}{x^2}=\frac 6{x^2}$$
as an alternative
$$f(x)=\frac{3x-6}{x}=3-\frac 6 x\implies f’(x)=\frac 6{x^2}$$
user
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in the alternate process you use, how o you go from the first step to the second step? and not only that, but why is it that you do that step? to simplify? – Pablo Aug 25 '18 at 09:12
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We simply have $$\frac{3x-6}{x}=\frac{3x}{x}-\frac{6}{x}=3-\frac{6}{x}$$ It simplify because we can use that the derivative of a constant is zero and that $(1/x)’=-1/x^2$. – user Aug 25 '18 at 09:45
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If one has to go by definition, then $$f'(a) = \lim _{x\to a} \left (\frac{3x-6}{x} -\frac{3a-6}{a}\right ) \cdot \frac{1}{x-a} = \lim _{x\to a} \frac{6x-6a}{ax(x-a)} = \frac{6}{a^2} $$
AlvinL
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