## LilySwan 6 months ago find dy/dx if (ax+cx^5)/kx^8

1. myininaya

a,c, and k are constants?

2. LilySwan

yes

3. micahwood50

Just use quotient rule.

4. LilySwan

got stucked at one point

5. micahwood50

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

6. LilySwan

why can we solve the numerator part first

7. micahwood50

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. LilySwan

for the upper part 5cx^4 +a

10. micahwood50

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

11. micahwood50

Can you find g'(x)?

12. LilySwan

for the lower part it will be 8kx^7

13. micahwood50

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. micahwood50

Can you do it? @LilySwan

15. LilySwan

trying wait

16. LilySwan

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. LilySwan

?

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. LilySwan

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

21. myininaya

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

22. LilySwan

yes

23. LilySwan

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