Derivative of 1/(4x^2 + 3x - x)

[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/(a+b) does not equal 1/a + 1/b !

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

Yikes!
If that were true, then this would follow:
[tex]\frac{1}{2} = \frac{1}{1+1} = \frac{1}{1} + \frac{1}{1} = 2[/tex]

 

[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 [itex](4x^2+2x)^{-1}[/itex] and use the regular chain rule.

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

[edit] fixed my tex tags