1. ## Derivatives

I just want someone to check my derivatives to see if I did them right:

$f(x) = \frac{(ax+b)}{(x-1)(x-4)}$

$f'(x) = \frac{(ax+b)(2x-5)-a(x-4)(x-1)}{((x-1)(x-4))^2}$

$f''(x) = \frac{(ax+b)}{-(2x-5)^3(x^2-5x+4)^2}$

2. This is assuming that $a$ and $b$ are constants?

First multiply out the bottom.

$\frac {(ax+b)}{(x-1)(x-4)} = \frac {ax+b}{x^2-5x+4}$

The quotient rule states that:

$\frac {d}{dx}(\frac {u}{v})=\frac {vu'-uv'}{v^2}$

So, by the quotient rule we have:

$\frac {(x^2-5x+4)(ax+b)'-(ax+b)(x^2-5x+4)'}{(x^2-5x+4)^2}$

$= \frac {(x^2-5x+4)(a)-(ax+b)(2x-5)}{(x^2-5x+4)^2}$

$= \frac {a(x-4)(x-1)-(ax+b)(2x-5)}{(x-4)^2(x-1)^2}$

Seems like you understand it, but remember the quotient rule:

(bottom*derivative of the top)-(top*derivative of the bottom) / (bottom squared)

3. Originally Posted by topher0805
$= \frac {a(x-4)(x-1)-(ax+b)(2x-5)}{(x-4)^2(x-1)^2}$
So that is the correct first derivative, right?

4. ***I basically need the second derivative so I can find the variables $a$ and $b$ given that the relative minimum is (2,-1). (Unless there is another way to solve the problem WITHOUT the second derivative.)

Sorry for the double post, but something doesn't seem right.

Here's my second derivative (too big to put into fraction form, I guess):

Numerator: $[(x-1)^2(x-4)^2][2ax-5a-4ax-2b-5a]$ $-[a(x-1)(x-4)-(ax+b)(2x-5)][(4x^3-66x)$ $-(30x^2+40)$

Denominator:
$(x-1)^4(x-4)^4$

5. Ok, I was assuming that a and b were constants. If they are variables that changes everything.

6. Originally Posted by topher0805
Ok, I was assuming that a and b were constants. If they are variables that changes everything.
No, they're constants. (Sorry about that. )

7. Originally Posted by Cursed
***I basically need the second derivative so I can find the variables $a$ and $b$ given that the relative minimum is (2,-1). (Unless there is another way to solve the problem WITHOUT the second derivative.)

[snip]
The point (2, -1) is on the curve.

f(2) = -1:

2 = 2a + b .... (1)

f'(2) = 0:

b = 0 .... (2)

Solve (1) and (2) simultaneously.

8. So if I did this correctly, my answers of $a=-1$ and $b=4$ are correct, right?

9. Originally Posted by Cursed
So if I did this correctly, my answers of $a=-1$ and $b=4$ are correct, right?
*Ahem* Take another look at equation (2) .......

10. Originally Posted by mr fantastic
*Ahem* Take another look at equation (2) .......
And you get equation (2) as follows:

0 = a(2 - 4)(2 - 1) - (2a + b)(4 - 5)

=> 0 = a(-2)(1) - (2a + b) (-1)

=> 0 = -2a + 2a - b

=> b = 0.