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How would I go at finding the derivative of:

$f(x)=\frac{3x-1}{x+{2}},x≠-2 \mathbb{}$

using first principles?

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    $\begingroup$ Welcome to Mathematics Stack Exchange. You could apply your definition of derivative. What is that? $\endgroup$ Jun 11 '19 at 22:37
  • $\begingroup$ that's all the information the question gives me. $\endgroup$
    – jakeymaths
    Jun 11 '19 at 22:40
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    $\begingroup$ Do you know what a derivative is? If not, you’re going to have a difficult time with this question $\endgroup$ Jun 11 '19 at 22:41
  • $\begingroup$ oh sorry, well I believe its the rate of change of a function? $\endgroup$
    – jakeymaths
    Jun 11 '19 at 22:43
  • $\begingroup$ @J.W. Tanner .. Ahh I didn't see. I will delay the answer. :) $\endgroup$
    – mm-crj
    Jun 11 '19 at 22:46
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For $x_0\ne2$ we have by definition of the derivative: \begin{align} f'(x_0) &= \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}\\ &= \lim_{x\to x_0} \left(\frac{3x-1}{x+2} - \frac{3x_0-1}{x_0+2}\right)\cdot \frac1{x-x_0}\\ &= \lim_{x\to x_0} \frac{(3x-1)(x_0+2)-(3x_0-1)(x+2)}{(x+2)(x_0+2)(x-x_0)}\\ &= \lim_{x\to x_0} \frac{7(x-x_0)}{(x+2)(x_0+2)(x-x_0)}\\ &= \lim_{x\to x_0} \frac7{(x+2)(x_0+2)}\\ &= \frac7{(x+2)^2}. \end{align}

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The definition of derivative, applied to this function, is $$ \lim_{h\to 0}\frac1{h} \left[ \frac{3(x+h)-1}{(x+h)+2}-\frac{3x-1}{x+2}\right] $$ Cross multiply to get $$ \lim_{h\to 0} \left[ \frac{(x+2)(3x+3h-1)-(x+h+2)(3x-1)}{h(x+h+2)(x+2)}\right]\\ = \lim_{h\to 0} \left[ \frac{3x^2+3hx-x+6x+6h-2-3x^2+x-3hx+h-6x+2}{h(x+h+2)(x+2)}\right]\\ = \lim_{h\to 0} \left[ \frac{3hx+6h-3hx+h}{h(x+h+2)(x+2)}\right]\\ = \lim_{h\to 0} \left[ \frac{3x+6-3x+1}{(x+h+2)(x+2)}\right] \\ = \lim_{h\to 0} \left[ \frac{7}{(x+h+2)(x+2)}\right] $$ Up to here we have not yet used the intention of taking $h \to 0$ but now we do, getting the answer $$ \frac7{(x+2)^2} $$

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This is quite straight forward actually. $f'(x)$ is defined according to the first principle as $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

\begin{align} &f(x)=\frac{3x-1}{x+2}\\ &f(x+h)=\frac{3x+3h-1}{x+h+2}\\ &\implies \frac{f(x+h)-f(x)}{h}=\frac{(3x-3h-1)(x+2)-(3x-1)(x+h+2)}{h}\\ &\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}= \lim_{h\to 0} \frac{7h}{h(x+h+2)(x+2)}\\ &=\frac{7}{(x+2)^2} \end{align}

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