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I can find this without using the limit definition (I think the formula is $\frac{f(x+h) - f(x)}{h}$

My first steps to solving are $f'(x) \lim\limits_{h\rightarrow 0} \frac{(x+h)^9 - (x+h)^7 - x^9 + x^7}{h}$

Thank you in advanced! This problem is making my head throb... and I already know the derivative is $9x^8 - 7x^6$

EDIT: Thank you everyone for the extra push! I was in the right direction, I just overcomplicated and did not realize that I could factor the h out of the problem after using the binomial theorem.

After plugging in the lim h->0 I had f'(x) - x^9+9x^8-7x^6-x^7-x^9+x^7 = 9x^8 - 7x^6

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    $\begingroup$ Have you tried using binomial expansion $\endgroup$ – user210387 May 31 '15 at 15:32
  • $\begingroup$ @Rememberme, may be you need to expand on your comment and post it as an answer. $\endgroup$ – abel May 31 '15 at 15:39
  • $\begingroup$ Well @abel many people have answered according to what I mean. So I think OP's demands have been fulfilled :) $\endgroup$ – user210387 May 31 '15 at 15:41
  • $\begingroup$ @Rememberme, i think you are just being lazy. $\endgroup$ – abel May 31 '15 at 15:44
  • $\begingroup$ Well @abel we have one answer with 6 upvotes and it seems its fair enough to go with the question So i feel its better that I dont write an answer unnecessarily $\endgroup$ – user210387 May 31 '15 at 15:46
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HINT:

$$(x+h)^9 - (x+h)^7 - x^9 + x^7$$ $$=\binom91x^8h+\binom92x^7h^2+O(h^3)-\left[\binom71x^6h+\binom72x^5h^2+O(h^3)\right]$$

As $h\to0,h\ne0$ so cancel $h$ safely.

In fact using Binomial series,

$(x+h)^n=x^n\left(1+\dfrac hx\right)^n=x^n\left(1+n\cdot\dfrac hx+O(h^2)\right)=x^n+nx^{n-1}h+O(h^2)$

$$\implies\dfrac{d(x^n)}{dx}=\lim_{h\to0}\dfrac{(x+h)^n-x^n}h=\lim_{h\to0}\dfrac{nx^{n-1}h+O(h^2)}h=nx^{n-1}$$

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You can also use

$\displaystyle f^{\prime}(x)=\lim_{t\to x}\frac{f(t)-f(x)}{t-x}=\lim_{t\to x}\frac{(t^9-t^7)-(x^9-x^7)}{t-x}=\lim_{t\to x}\frac{(t^9-x^9)-(t^7-x^7)}{t-x}$

$=\displaystyle\lim_{t\to x}\frac{(t-x)[t^8+t^7 x+t^6 x^2+\cdots+x^8]-(t-x)[t^6+t^5 x+t^4 x^2+\cdots+x^6]}{t-x}$

$=\displaystyle\lim_{t\to x}\left([t^8+t^7 x+t^6 x^2+\cdots+x^8]-[t^6+t^5 x+t^4 x^2+\cdots+x^6]\right)=9x^8-7x^6$

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Good start.

From here, you can expand the binomials, such as $(x+h)^9$ without as much effort as you might think, by using the binomial theorem.

See if you can do that, then start factoring the numerator.

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