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

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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)
why in the second step the exponent became 2...did you subtract
dividing x by x^3

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Other answers:

ok thanks
welcum ! :-)
is your expression on right hand side 5x/(x^3 -4) ??
well i thought 4 is not included , and acc I have diffrenctd , whereas had it been with x^3 , ans would have been diff
now if its included , quotient rule has to be applied
in the denominator it is....x^(3) - 4
oops dn my ans is wrong ! wait I'll explain you this one
\[ f(x) = \frac {5x}{x^3 -4} \implies f'(x) = \frac{(x^3 -4)\times 5 -5x \times(3x)}{(x^3 -4)^2} \]
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 ! :-)
fool for math, why isnt it 3x^2 in the numerator
Lol, because It was a typo.
\[ f(x) = \frac {5x}{x^3 -4} \implies f'(x) \] \[= \frac{(x^3 -4)\times 5 -5x \times(3x^2)}{(x^3 -4)^2} = -\frac{10 \left(2+x^3\right)}{\left(-4+x^3\right)^2} \]
Do check my algebra, don't ever forget that I am a Fool ;) :P
looks right...-10x^3 - 20/ (x^3 - 4)^2
yea thats right ! :-)
Wow! Congratz too me :D Yay!!!
just strict to the formula , and you ll never be wrong !
She mean stick.
oops ! typing error , stick ! :p
lol, it happens to non-fools too :D
LOL ! :D

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