Derivatives - Quotient Rule

Alpha6

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I was wondering if I did this right, and if not what to correct please.

I am looking for the derivative of z = 4-3x / 3x2

I understand the quotient rule as: Derivative of numerator * denominator Minus numerator times the Derivative of denominator all over Denominator squared.

So I did (4 -3X)' * (3X2 + X) - (4-3X) * (3X2+ X)' / (3X2+X)2


(3) * (3X2 + X) - (4 - 3X) * (6X+1) / (3X2+X)2



9X2 +3X - 18X2 -27X - 4 / (3X2+X)2





-9X2 -24X -4 / 9X4 +X2

 
I was wondering if I did this right, and if not what to correct please.

I am looking for the derivative of z = 4-3x / 3x2
What you have posted means this:

. . . . .\(\displaystyle z\, =\, 4\, -\, \dfrac{3x}{3x^2}\, =\, 4\, -\, \dfrac{1}{x}\)

However, I suspect that you meant "z = (4 - 3x)/(3x^2)", or:

. . . . .\(\displaystyle z\, =\, \dfrac{4\, -\, 3x}{3x^2}\)

Did I guess correctly?

So I did (4 -3X)' * (3X2 + X) - (4-3X) * (3X2+ X)' / (3X2+X)2
How did the denominator, 3x^2, turn into 3x^2 + x? Where did the "+x" come from? ;)
 
Assuming that the problem is to differentiate \(\displaystyle \frac{4- 3x}{3x^2}\) then, yes, you can use the quotient rule.

The quotient rule says that \(\displaystyle \left(\frac{u(x)}{v(x)}\right)'= \frac{u'(x)v(x)- u(x)v'(x)}{u^2(x)}\).

Here, that is \(\displaystyle \frac{(4- 3x)'(3x^2)- (4- 3x)(3x^2)'}{9x^4}\). What are \(\displaystyle (4- 3x)'\) and \(\displaystyle (3x^2)'\) ?
(In your first post, you have "3" as the derivative of 4- 3x. That is wrong.)

In this particular problem, since the denominator, \(\displaystyle 3x^2\) is a single term, you can write it as
\(\displaystyle (4/3)x^{-2}- x^{-1}\) and differentiate that. Do it both ways as a check.
 
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mea culpa

I made a mistake posting the original equation, I should proofread it.

This is what it should be: z = (4-3x) / 3x2 +x

That's where the "+x" came from sorry!

So I'll do it as...

(4-3x)' * (3x2+X) - (4-3x) * (3x2 +x)'



(-3) * (3x2+X) - (4-3x) * 6x



-9x2 -3x - ( 24x -18x2)



-9x2 -3x -24x + 18x2


9x2 - 27x
 
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