## anonymous 2 years ago find dy/dx if (ax+cx^5)/kx^8

1. myininaya

a,c, and k are constants?

2. anonymous

yes

3. anonymous

Just use quotient rule.

4. anonymous

got stucked at one point

5. anonymous

$$\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$$

6. anonymous

why can we solve the numerator part first

7. anonymous

treat constant as if they are just "number", and not variable.

8. myininaya

@LilySwan Did you have a problem with finding derivative of the top or the bottom? That is what you need to plug into @micahwood50 's formula. what do you mean solve the numerator? You shouldn't be solving. You should be differentiating it. What is the derivative of the top part of your fraction?

9. anonymous

for the upper part 5cx^4 +a

10. anonymous

Right f'(x) = a+5cx^4

11. anonymous

Can you find g'(x)?

12. anonymous

for the lower part it will be 8kx^7

13. anonymous

Yep. So you have f(x) = ax+cx^5, f'(x) = a+5cx^4, g(x) = kx^8, g'(x) = 8kx^7 Just plug these in formula given above then simplify.

14. anonymous

Can you do it? @LilySwan

15. anonymous

trying wait

16. anonymous

i was told to use this formula $\frac{ d }{ dx }(\frac{ u}{ v }) =\frac{ (v \frac{ du }{ dx })-(u \frac{ dv }{ dx }) }{ v^2 }$

17. myininaya

same formula different notation

18. anonymous

?

19. myininaya

$\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$ if you want you can replace the f's with u's the g's with v's the ' things with d/dx and you have the same formula

20. anonymous

$5cx^4+a.kx^8 - ax+cx^5. 8kx^7/(kx^8)^2$

21. myininaya

$\frac{(5cx^4+a)(kx^8)-(ax+cx^5)(8kx^7)}{(kx^8)^2}$ i think this is what you mean

22. anonymous

yes

23. anonymous

what should i be doing the next

24. myininaya

guess you could multiply things out on top and also use law of exponents on the bottom to simplify the bottom