Stuck on derivative.

**el123** · August 11th 2009, 10:16 PM

I have equation

$y=\frac{2x-5}{3x+2}$

Please check my working.(I started expanding the right hand side of the numerator first , i hope you can follow.

$=\frac{[(3x+2)^2(2)]-[(2x-5)[(3x+2)(3x+2)]}{[(3x+2)^2]^2}$

$=\frac{[(3x+2)^2(2)]-[(2x-5)(9x^2+6x+6x+4)]}{[3x+2)^2]^2}$

$=\frac{[(3x+2)^2(2)]-[(2x-5)(9x^2+12x)]}{[(3x+2)^2]^2}$

$=\frac{[(3x+2)^2(2)]-[(2x-5)(18x+12)]}{[(3x+2)^2]^2}$

$=\frac{[((3x+2)(3x+2)(2)]-(36x+24x-90x-60)}{[(3x+2)(3x+2)(3x+2)(3x+2)]}$

$=\frac{((3x+2)(3x+2)(2)-36x-24x+90x+60} {(3x+2)(3x+2)(3x+2)(3x+2)}$

$=\frac{(9x^2+12x+4)(2)+30x+60}{54x+78x+24}$
$=\frac{(18x^2+24x+8)+30x+60}{132x+24}$
$=\frac{18x^2+54x+68}{132x+24}$

ok i know thats alot to read but if anyone is able to let me know if ive gone wrong anywhere it would be much appreciated.

**Chris L T521** · August 11th 2009, 10:59 PM

Originally Posted by el123

[snip]

I have equation

$y=\frac{2x-5}{3x+2}$

Please check my working.(I started expanding the right hand side of the numerator first , i hope you can follow.

$=\frac{[(3x+2)^{{\color{red}2}}(2)]-[(2x-5)[{\color{blue}(3x+2)}{\color{red}(3x+2)}]}{[(3x+2)^2]^{{\color{red}2}}}$

[/snip]

The terms in red aren't needed, and the term in blue needs to be differentiated. That will give you the answer you're looking for.

**matheagle** · August 11th 2009, 11:20 PM

I would divide first to make the calculus a lot easier.
That way you can avoid the quotient rule.

**el123** · August 11th 2009, 11:41 PM

Darn i just realised the very first equation was

$\frac{2x-5}{(3x+2)^2}$

does it make sense now or still not right?

Woulld you be able to explain the steps of how i should begin to diffrentiate it?

**skeeter** · August 12th 2009, 04:41 AM

Originally Posted by el123

Damn i just realised the very first equation was

$\frac{2x-5}{(3x+2)^2}$

does it make sense now or still not right?

Woulld you be able to explain the steps of how i should begin to diffrentiate it?

seems like I've seen this question before ...

$\frac{d}{dx}\left[\frac{2x-5}{(3x+2)^2}\right]$

$\frac{(3x+2)^2(2) - (2x-5) \cdot 6(3x+2)}{(3x+2)^4}$

$\frac{34-6x}{(3x+2)^3}$

you need to learn about the chain rule for derivatives so that you may easily find derivatives of functions like $(3x+2)^2$

**mr fantastic** · August 12th 2009, 04:49 AM

Originally Posted by el123

Darn i just realised the very first equation was

$\frac{2x-5}{(3x+2)^2}$

does it make sense now or still not right?

Woulld you be able to explain the steps of how i should begin to diffrentiate it?

Logarithmic differentiation can also be used: Logarithmic Differentiation

**el123** · August 12th 2009, 02:15 PM

Originally Posted by skeeter

seems like I've seen this question before ...

$\frac{d}{dx}\left[\frac{2x-5}{(3x+2)^2}\right]$

$\frac{(3x+2)^2(2) - (2x-5) \cdot 6(3x+2)}{(3x+2)^4}$

$\frac{34-6x}{(3x+2)^3}$

you need to learn about the chain rule for derivatives so that you may easily find derivatives of functions like $(3x+2)^2$

ok i can see how you get the last part of the numerator and ive been learning the chain rul ebut i cant seem to see how you expanded that to get 34-6x/3x+2^3

Could you explain please. I am so confused. I keep getting answers that are way off the mark

**skeeter** · August 12th 2009, 02:42 PM

note the common factor (3x+2) in both terms of the numerator ...

$\frac{\textcolor{red}{(3x+2)^2}(2) - (2x-5) \cdot 6\textcolor{red}{(3x+2)}}{(3x+2)^4}$

factor out one (3x+2) from both terms ...

$\frac{\textcolor{red}{(3x+2)}[(3x+2)(2) - (2x-5) \cdot 6]}{(3x+2)^4}$

cancel the common factor with one in the denominator ...

$\frac{(3x+2)(2) - (2x-5) \cdot 6}{(3x+2)^3}$

clean up the like terms ...

$\frac{6x+4 - 12x+30}{(3x+2)^3}$

$\frac{34-6x}{(3x+2)^3}$

the hardest thing about calculus is the algebra.

**tom@ballooncalculus** · August 12th 2009, 11:57 PM

... perhaps matheagle is right to suggest avoiding the quotient rule, which often raises the power in the denominator unnecessarily. Another way of avoiding it is to use a chain rule...

inside a product rule...

giving...

or

...where the simplest common denominator is formed directly.

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