## anonymous one year ago Is this correct? (Spoiler: It's derivatives) http://i.imgur.com/MBKQYup.png

1. phi

what is the derivative of $7 x^{-1}$ ?

2. phi

using the power rule d/dx x^n = n x^(n-1)

3. anonymous

I wish I could tell you - the content doesn't cover that. It only gives me f(x) and ask me to find it.

4. phi

The rule is posted above. But here it is again $\frac{d}{dx} x^n = n x^{n-1}$

5. phi

6. anonymous

What would variable "d" be then?

7. anonymous

The $$d$$ is notation for an operator. Just like $$+$$, it doesn't have a value

8. phi

d/dx is how you say "derivative with respect to x"

9. phi

the power rule for finding a derivative is one of the first things you learn in differential calculus. If this is mysterious, you need a refresher, because it is too much material to review here.

10. anonymous

I'm in Precalc, so we're probably never gonna cover the power rule. This is simply using the difference quotient.

11. phi

you mean using $f'(x) = \lim_{h\rightarrow0}\frac{ f(x+h) - f(x)}{(x+h)-x}$?

12. anonymous

Yep, we're doing that to solve for the posted equation.

13. phi

in that case write $f'(x) = \lim_{h\rightarrow0}\frac{\frac{7}{x+h} - \frac{7}{x} }{h}$

14. anonymous

Alright, sorry. Internet kinda cut out. I know we gotta find a common denominator between the two fractions.

15. anonymous

(which would be x for the first fraction, (x+h) for the second if I'm correct in saying that)

16. phi

or $f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right)$

17. phi

multiply the first fraction by x/x and the second by (x+h)/(x+h)

18. anonymous

$\frac{ x }{ x^2 +xh} - \frac{ x+h }{ x^2+xh }$ Right?

19. phi

ok. now combine the top, and put the result over the common denominator

20. anonymous

$\frac{ h }{ x^2+xh }$ Alrightie.

21. phi

the top is x-(x+h) = x-x-h

22. phi

$f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x} \right) \\ f'(x) = 7\lim_{h\rightarrow0} \frac{1}{h}\left(\frac{-h}{x^2+xh} \right)$

23. anonymous

Alright, I see.

24. phi

notice h/h divides out and you have $f'(x) = 7\lim_{h\rightarrow0} \left(\frac{-1}{x^2+xh} \right)$

25. phi

now let h "go to zero". the expression approaches -1/(x^2) you get $f'(x) = 7 \cdot \frac{-1}{x^2}$ evaluate that at x=1

26. anonymous

So from here I'd replace x^2 with 1?

27. phi

you replace x with 1 so you get 1^2 or 1*1 in the denominator.

28. anonymous

Which gives the end result of -7?

29. phi

yes

30. anonymous

Alright, thank you so much for your help.

31. phi

yw

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