Question

if y= 5x/x^3 -4, then dy/dx =?

Answer 1

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)

Answer 2

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

Answer 3

dividing x by x^3

Answer 4

ok thanks

Answer 5

welcum ! :-)

Answer 6

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

Answer 7

yea

Answer 8

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

Answer 9

now if its included , quotient rule has to be applied

Answer 10

wait

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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 ! :-)

Answer 15

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

Answer 16

Lol, because It was a typo.

Answer 17

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

Answer 18

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

Answer 19

lol

Answer 20

;)

Answer 21

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

Answer 22

yea thats right ! :-)

Answer 23

Wow! Congratz too me :D Yay!!!

Answer 24

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

Answer 25

She mean stick.

Answer 26

oops ! typing error , stick ! :p

Answer 27

lol, it happens to non-fools too :D

Answer 28

lol

Answer 29

LOL ! :D