## Jinnie 3 years ago if y= 5x/x^3 -4, then dy/dx =?

1. anjali_pant

y=5x/x^3 -4 y=5/x^2-4 y=5*[x^(-2)}-4 dy/dx=5*(-2) [ x^(-2-1)]-0 dy/dx=-10*x^(-3)

2. Jinnie

why in the second step the exponent became 2...did you subtract

3. anjali_pant

dividing x by x^3

4. Jinnie

ok thanks

5. anjali_pant

welcum ! :-)

6. experimentX

is your expression on right hand side 5x/(x^3 -4) ??

7. Jinnie

yea

8. anjali_pant

well i thought 4 is not included , and acc I have diffrenctd , whereas had it been with x^3 , ans would have been diff

9. anjali_pant

now if its included , quotient rule has to be applied

10. Jinnie

wait

11. Jinnie

in the denominator it is....x^(3) - 4

12. anjali_pant

oops dn my ans is wrong ! wait I'll explain you this one

13. FoolForMath

$f(x) = \frac {5x}{x^3 -4} \implies f'(x) = \frac{(x^3 -4)\times 5 -5x \times(3x)}{(x^3 -4)^2}$

14. anjali_pant

quotient rule : {(dy/dx numerator)*denominator-[(dy/dx denominator)*numerator]} / denominator^2 try solving with this , if you get stuck , letme know ! btw FFM has done a great job already , so you wont find any pro ! :-)

15. Jinnie

fool for math, why isnt it 3x^2 in the numerator

16. FoolForMath

Lol, because It was a typo.

17. FoolForMath

$f(x) = \frac {5x}{x^3 -4} \implies f'(x)$ $= \frac{(x^3 -4)\times 5 -5x \times(3x^2)}{(x^3 -4)^2} = -\frac{10 \left(2+x^3\right)}{\left(-4+x^3\right)^2}$

18. FoolForMath

Do check my algebra, don't ever forget that I am a Fool ;) :P

19. Jinnie

lol

20. FoolForMath

;)

21. Jinnie

looks right...-10x^3 - 20/ (x^3 - 4)^2

22. anjali_pant

yea thats right ! :-)

23. FoolForMath

Wow! Congratz too me :D Yay!!!

24. anjali_pant

just strict to the formula , and you ll never be wrong !

25. FoolForMath

She mean stick.

26. anjali_pant

oops ! typing error , stick ! :p

27. FoolForMath

lol, it happens to non-fools too :D

28. Jinnie

lol

29. anjali_pant

LOL ! :D