# Thread: simplifying a quotient rule question of f"

1. ## simplifying a quotient rule question of f"

Find f '(x) and f ''(x) of $\displaystyle f(x)= (x^2)/(1+4x)$

I found f':
f'= $\displaystyle (2x+4x^2)/(1+4x)^2$.

for f"
$\displaystyle [(2+8x)(1+4x)^2 - 2(1+4x)(4)]/[(1+4x)^2]^2$

*I know this is correct, but I have to input it in an online grader so it has to be in it's most simplified form.*
so...
factoring a $\displaystyle (1+4x)^2$=
$\displaystyle (1+4x)[(2+8x)(1+4x)-8]/[(1+4x)^4]$
$\displaystyle (1+4x)[(2+8x+8x+32x^2)-8]/[(1+4x)^4]$
$\displaystyle (1+4x)[(2+8x+8x+32x^2)-8/[(1+4x)^4]$
$\displaystyle (1+4x)[(16x+32x^2-6/[(1+4x)^4]$
factoring out a 2:
$\displaystyle (2+8x)[(16x^2+8x-3)]/[(1+4x)^4]$ *was marked wrong*

Any arithmetic/calculus errors or can it be simplified further?

2. Originally Posted by hazecraze
Find f '(x) and f ''(x) of f(x)= (x^2)/(1+4x)

I found f':
f'= (2x+4x^2)/(1+4x)^2.

for f"
[(2+8x)(1+4x)^2 - 2(1+4x)(4)]/[(1+4x)^2]^2

*I know this is correct, but I have to input it in an online grader so it has to be in it's most simplified form.*
so...
factoring a (1+4x)^2=
(1+4x)[(2+8x)(1+4x)-8]/[(1+4x)^4]
(1+4x)[(2+8x+8x+32x^2)-8]/[(1+4x)^4]
(1+4x)[(2+8x+8x+32x^2)-8]/[(1+4x)^4]
(1+4x)[(16x+32x^2-6]/[(1+4x)^4]
factoring out a 2:
(2+8x)[(16x^2+8x-3)]/[(1+4x)^4] *was marked wrong*

Any arithmetic/calculus errors or can it be simplified further?
what's $\displaystyle \frac {2+8x} {1 + 4x}$?

3. 2, so 2(16x^2+8x-3)/(1+4x)^3?

4. Have you checked to make sure that $\displaystyle 16x^2+ 8x- 3$ doesn't have a factor of 4x+ 1?

5. Originally Posted by hazecraze
2, so 2(16x^2+8x-3)/(1+4x)^3?
You can also factorise the quadratic. Go on, give it a go.

6. Originally Posted by HallsofIvy
Have you checked to make sure that $\displaystyle 16x^2+ 8x- 3$ doesn't have a factor of 4x+ 1?
Shouldn't it be 4x- 1? (4x- 1)(4x+3)=$\displaystyle 16x^2+ 12x-4x- 3$=$\displaystyle 16x^2+ 8x- 3$. That messes up the canceling of the denominator though, which I can see now is where we were going.

7. Originally Posted by hazecraze
Find f '(x) and f ''(x) of f(x)= (x^2)/(1+4x)

I found f':
f'= (2x+4x^2)/(1+4x)^2.

for f"
[(2+8x)(1+4x)^2 - 2(1+4x)(4)]/[(1+4x)^2]^2

*I know this is correct, but I have to input it in an online grader so it has to be in it's most simplified form.*
I'm not sure it is.

Let's put math delimiters round it to start with so we can see the silly thing:

$\displaystyle f(x)= \frac {x^2} {1+4x}$

$\displaystyle f'(x) = \frac {2x+4x^2} {(1+4x)^2}$, yep, I'll agree there ...

Then you got your quotient rule $\displaystyle d \left({\frac u v}\right) = \frac {v du - u dv} {v^2}$

$\displaystyle u = 2x+4x^2, v = (1+4x)^2$ so $\displaystyle du = 2 + 8x, dv = 2 (1+4x) 4 = 8 (1+4x)$.

So $\displaystyle f''(x) = \frac {(1+4x)^2 (2 + 8x) - 8 (2x+4x^2) (1+4x)} {(1+4x)^4}$

I'll leave you to simplify it, I've got things to do today.

8. Doh! Don't know how I missed that. So, $\displaystyle \frac {2} {1+4x^3}$. I figured they'd give us something easy to simplify.

9. Originally Posted by hazecraze
Doh! Don't know how I missed that. So, $\displaystyle \frac {2} {1+4x^3}$. I figured they'd give us something easy to simplify.
You're not the only one to miss it - we all did! (to start with, at least.)