derivative

[UrbanXrisis]

what is the derivative of 1/(4x^2+3x-x)

I got 1/4x is that correct?

[Doc Al]

How did you get that?

[UrbanXrisis]

1/(4x^2+3x-x) =
(1/4x^2)+(1/3x)-(1/x)=
((4x^2)^-1 )+ ((3x)^-1) - ((x)^-1))=
4x+3-1/x
derivative = 4x^-1 = 1/4x

[Ebolamonk3y]

Wrong.

The answer is -(4x+1)/((2x^2)*(2x+1)^2)

[Doc Al]

1/(4x^2+3x-x) =
(1/4x^2)+(1/3x)-(1/x)=
Yikes! :eek:
If that were true, then this would follow:
\frac{1}{2} = \frac{1}{1+1} = \frac{1}{1} + \frac{1}{1} = 2

[UrbanXrisis]

oh wow, what was I thinking??!?!? whooops. How would I solve this then?

[faust9]

You could use the quotient rule, or simply rewrite the equation as (4x^2+2x)^{-1} and use the regular chain rule.

PS: \frac{1}{A+B+C} does not equal \frac{1}{A}+\frac{1}{B}+\frac{1}{C} ; however, \frac{A+B+C}{D} does equal \frac{A}{D}+\frac{B}{D}+\frac{C}{D}. See the difference? You cant simply split a denominator but you can split up a numerator. Use partial fractions to split the denominator.

[edit] fixed my tex tags