Derivatives

[Kondwani Hauya]

Differentiate the function below with respect to x
y = (3x^2 + 2x + 1)/(4x^2 - 5)

Below is the solution to the above problem. May somebody check it?

- (21x^2 + 15x + 5)/(4x^2 - 5x)^2

I have deliberately chose not to show every step of the solution because my focus is on whether I got it right or wrong

 

[pka]

Differentiate the function below with respect to x
y = (3x^2 + 2x + 1)/(4x^2 - 5)
Below is the solution to the above problem. May somebody check it?
- (21x^2 + 15x + 5)/(4x^2 - 5x)^2
I have deliberately chose not to show every step of the solution because my focus is on whether I got it right or wrong

The denominator is correct. Now work out the numerator:
\(\displaystyle (6x+2)(4x^2-5)-(3x^2+2x+1)(8x)\)

 

[HallsofIvy]

No, the denominator is NOT correct. You have, as the denominator for the derivative, \(\displaystyle (4x^2- 5x)^2\) but the denominator in the original function is \(\displaystyle 4x^2- 5\). The denominator for the derivative should be \(\displaystyle (4x^2- 5)^2\). And the numerator should be \(\displaystyle (3x^2+ 2x+ 1)'(4x^2- 5)+ (3x^2+ 2x+1)(4x^2- 5)'

= (6x+ 2)(4x^2- 5)+ (3x^2+ 2x+ 1)(8x)

= 24x^3+ 8x^2- 30x- 10+ 24x^3+ 16x^2+ 8x= 48x^3+ 24x^3- 22x- 10$.\)

 

[Kondwani Hauya]

I regret to realise the typing error. The problem is supposed to be expressed as (3x^2+2x+1)/(4x^2-5x) . That's why my solution seemed to be incorrect. I humbly ask for your cooperation once again. Thanks

 

[JeffM]

I get a somewhat different numerator but your denominator.

 

[Kondwani Hauya]

I get a somewhat different numerator but your denominator.

May you show me the numerator. Maybe the difference arises due to arithmetic error

 

[JeffM]

It is always possible that it is I who have made an error

[MATH]f(x) = \dfrac{u(x)}{v(x)}, \text { where } u(x) = 3x^2 + 2x + 1 \text { and } v(x) = (4x^2 - 5x).[/MATH]
[MATH]\therefore f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{\{(v(x)\}^2} = \dfrac{u'(x)v(x) - u(x)v'(x)}{(4x^2 - 5x)^2}.[/MATH]
[MATH]u'(x) = 6x + 2 \text { and } v'(x) = 8x - 5.[/MATH]
[MATH]\therefore f'(x) = \dfrac{(4x^2 - 5x)(6x + 2) - \{(3x^2 + 2x + 1)(8x - 5)\}}{(4x^2 - 5x)^2} =[/MATH]
[MATH]\dfrac{24x^3 + 8x^2 - 30x^2 - 10x - (24x^3 - 15x^2 + 16x^2 - 10x + 8x - 5) }{(4x^2 - 5x)^2} =[/MATH]
[MATH]\dfrac{24x^3 - 22x^2 - 10x - (24x^3 + x^2 - 2x - 5) }{(4x^2 - 5x)^2} =[/MATH]
[MATH]\dfrac{24x^3 - 22x^2 - 10x - 24x^3 - x^2 + 2x + 5 }{(4x^2 - 5x)^2} =[/MATH]
[MATH]\dfrac{-\ 23x^2 - 8x + 5 }{(4x^2 - 5x)^2}.[/MATH]
You can of course expand the denominator, but I see no reason to do so.