How would I find the derivative of $f(x)=2x^{2}+2x+1?$ I know that $f'(x)=4x+2$ using the power rule, but but how should I solve it using difference quotient, can someone please show the step by step procedure? Thank you so much!
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$\begingroup$computing the difference quotient we obtain $$\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)^2+2(x+h)+1-2x^2-2x-1}{h}$$ simplify it and compute the limit for $$h\to0$$
$\endgroup$ $\begingroup$It actually doesn't cost much more to draw the general formula than to process a particular example.
Indeed, it is no big deal to establish
$\dfrac{(x+h)^n-x^n}h=nx_{n-1}+h^2P(x,h)$ where $P$ is a polynomial (by induction or by the Binomial theorem),
$\dfrac{(f(x+h)+g(x+h))-(f(x)+g(x))}h=\dfrac{f(x+h)-f(x)}h+\dfrac{g(x+h)-g(x)}h$,
$\dfrac{(cf(x+h))-(cf(x))}h=c\dfrac{f(x+h)-f(x)}h$.
Taking the limits $h\to0$ (provided they exist, which they do) and combining, you get the formulas for powers and for polynomials.
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