# Derivative question using the Quotient rule

• Mar 17th 2007, 08:18 AM
kcsteven
Derivative question using the Quotient rule
Ithe problem is as follows: d/dx[1-5x/1+5x] I need to find f'(x) and f''(x)
I found f'(x) but the second derivative is giving me problems. Here is what I did for both: d/dx[1-5x/1+5x] = (1+5x) d/dx [1-5x] - (1-5x) d/dx [1+5x] and all this is divided by (1+5x)^2
= (1+5x)(-5) - (1-5x)(5)/(1+5x)^2 =
-5 -25x - 5 + 25x/(1+5x)^2 =
-10/(1-5x) Now I tried to get f''(x) unsuccessfully I might add.
(1+10x+25x^2) d/dx [-10] - (-10) d/dx [1+10x+25x^2] all this is divided by
(1+10x+25x^2)^2 =
1+10x+25x^2 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
1+5010x+25x^2/(1+10x+25x^2)^2

Please let me know where I went off track.
THANK YOU!
Keith
• Mar 17th 2007, 08:36 AM
earboth
Quote:

Originally Posted by kcsteven
Ithe problem is as follows: d/dx[1-5x/1+5x] I need to find f'(x) and f''(x)
I found f'(x) but the second derivative is giving me problems. Here is what I did for both: d/dx[1-5x/1+5x] = (1+5x) d/dx [1-5x] - (1-5x) d/dx [1+5x] and all this is divided by (1+5x)^2
= (1+5x)(-5) - (1-5x)(5)/(1+5x)^2 =
-5 -25x - 5 + 25x/(1+5x)^2

Hello, Keith,

all calculations are correct but the next step:
= -10/(1+5x)^2

To get f''(x) it would be better to write f'(x) = (-10)*(1+5x)^(-2)
Now you only have to use the chain rule:
f''(x) = (-10)*(-2)*(1+5x)^(-3)*(5) = (100)/((1+5x)^3)

EB
• Mar 17th 2007, 08:57 AM
Pajkaj
Quote:

Originally Posted by kcsteven
Ithe problem is as follows: d/dx[1-5x/1+5x] I need to find f'(x) and f''(x)
I found f'(x) but the second derivative is giving me problems. Here is what I did for both: d/dx[1-5x/1+5x] = (1+5x) d/dx [1-5x] - (1-5x) d/dx [1+5x] and all this is divided by (1+5x)^2
= (1+5x)(-5) - (1-5x)(5)/(1+5x)^2 =
-5 -25x - 5 + 25x/(1+5x)^2 =
-10/(1-5x) Now I tried to get f''(x) unsuccessfully I might add.
(1+10x+25x^2) d/dx [-10] - (-10) d/dx [1+10x+25x^2] all this is divided by
(1+10x+25x^2)^2 =
1+10x+25x^2 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
1+5010x+25x^2/(1+10x+25x^2)^2

Please let me know where I went off track.
THANK YOU!
Keith

Hello Keith!

First, I know their was already a post submitted since I've been writing this, but I figured some teachers want you specifically to use the quotient rule over the power rule. Which sucks, because I love the power rule. So using only the quotient rule, your (small) mistakes are below:

To begin, I, without looking at your work, equated f'(x) and got the same answer as you:
f'(x) = -10/[(1+5x)^2]

I think I've found your problem. When you are taking this part of your f''(x):

(1+10x+25x^2) d/dx [-10]

You say that the above argument equals (1+10x+25x^2). But d/dx of -10, or any constant, is 0, not 1. So:

(1+10x+25x^2) d/dx [-10] equals 0 instead of (1+10x+25x^2)

So looking at your work again:

(1+10x+25x^2) d/dx [-10] - (-10) d/dx [1+10x+25x^2] all this is divided by
(1+10x+25x^2)^2 =
1+10x+25x^2 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
1+5010x+25x^2/(1+10x+25x^2)^2

The result would change to:

0 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
5000x/(1+10x+25x^2)^2

The other problem lies at this point in your work:
(-10) d/dx [1+10x+25x^2] = (-10)(10)(50x)
(-10) d/dx [1+10x+25x^2] should equal:
(-10) * (50x+10) which equals: -500x -100

So we now have:

0 - (-500x - 100) / (1+10x+25x^2)^2=
(500x + 100) / (1+10x+25x^2)^2 =
(500x + 100) / (1+5x)^4

Now, if you really want to take it further,
(1 +5x)^4 can be expanded to:
625x^4 + 500x^3 + 150x^2 + 20x + 1

So an optional f''(x) answer is:
(500x+100)/(625x^4 + 500x^3 + 150x^2 + 20x + 1)
But if it were me I'd leave the denominator unexpanded.

f'(x) = (-10) / (1 + 5x)^2
f''(x) = (500x + 100) / (1 + 5x)^4

Hope this helps!

P.S.

To get the above answer, we factor (100) out of the numerator:
(500x +100) / (1 + 5x)^4 = (100)(1 + 5x) / (1 + 5x)^4
The (1 + 5x)'s cancel out, giving us:
100 / (1 + 5x)^3, the answer from the above post.
• Mar 17th 2007, 09:05 AM
Soroban
Hello, Keith!

Quote:

f(x) .= .(1-5x)(1+5x) . Find f'(x) and f''(x)

Your first derivative is correct: .f'(x) .= .10/(1 + 5x)²
. . but why multiply it out?