# Thread: Stuck on derivative.

1. ## Stuck on derivative.

I have equation

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

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

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

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

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

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

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

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

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

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.

2. Originally Posted by el123

[snip]

I have equation

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

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

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

[/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.

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

4. Darn i just realised the very first equation was

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

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?

5. Originally Posted by el123
Damn i just realised the very first equation was

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

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 ...

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

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

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

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

6. Originally Posted by el123
Darn i just realised the very first equation was

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

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

7. Originally Posted by skeeter
seems like I've seen this question before ...

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

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

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

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

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

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

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

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

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

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

clean up the like terms ...

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

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

the hardest thing about calculus is the algebra.

9. ... 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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