Differentiate $\frac{3x-6}{x}$

Question

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?

Answer 1

By quotient rule

$$f’(x)=\frac{3x-(3x-6)}{x^2}=\frac 6{x^2}$$

as an alternative

$$f(x)=\frac{3x-6}{x}=3-\frac 6 x\implies f’(x)=\frac 6{x^2}$$

Answer 2

If one has to go by definition, then $$f'(a) = \lim _{x\to a} \left (\frac{3x-6}{x} -\frac{3a-6}{a}\right ) \cdot \frac{1}{x-a} = \lim _{x\to a} \frac{6x-6a}{ax(x-a)} = \frac{6}{a^2} $$

Answer 3

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 $(*)$: $$\left(\frac{3x-6}{x}\right)'=\frac{(3x-6)'x-(3x-6)x'}{x^2}=\frac{3x-(3x-6)}{x^2}=\frac{6}{x^2}$$

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$(*)$ 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 $\frac{f'(x)}{g'(x)}$...!