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

  • 4
    $\begingroup$ 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


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

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

In fact using Binomial series,

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



You can also use

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

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

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.