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