# Very cool

• Dec 20th 2009, 09:47 AM
dapore
Very cool
D (x) = (x + 1) (x + 2) ^ 2 (x +3) ^ 3 (x + 4) ^ 4 (x + 5)

Find the derivative at -1
• Dec 20th 2009, 09:51 AM
e^(i*pi)
Quote:

Originally Posted by dapore
D (x) = (x + 1) (x + 2) ^ 2 (x +3) ^ 3 (x + 4) ^ 4 (x + 5)

Find the derivative at -1

Use log differentiation

$ln[D(x)] = ln[(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]$ $= ln(x+1) + 2ln(x+2) + 3ln(x+3) + 4ln(x+4) + ln(x+5)$

Recall that due the chain rule: $\frac{d}{dx}ln[f(x)] = \frac{f'(x)}{f(x)}$

Actually, looking at it I don't think it is differentiable at x=-1 because of the 1/(x+1) term
• Dec 20th 2009, 10:04 AM
skeeter
Quote:

Originally Posted by dapore
D (x) = (x + 1) (x + 2) ^ 2 (x +3) ^ 3 (x + 4) ^ 4 (x + 5)

Find the derivative at -1

$D(x) = f(x) \cdot g(x) \cdot h(x) \cdot p(x) \cdot q(x)
$

$D'(x) = f'(x) \cdot g(x) \cdot h(x) \cdot p(x) \cdot q(x)$ $+ f(x) \cdot g'(x) \cdot h(x) \cdot p(x) \cdot q(x)$ $+ f(x) \cdot g(x) \cdot h'(x) \cdot p(x) \cdot q(x)$ $+ f(x) \cdot g(x) \cdot h(x) \cdot p'(x) \cdot q(x)$ $+ f(x) \cdot g(x) \cdot h(x) \cdot p(x) \cdot q'(x)$

since $f(x) = (x+1)$ , and $f(-1) = 0$ ...

$D'(-1) = (1)(1^2)(2^3)(3^4)(4)$
• Dec 20th 2009, 11:23 AM
Jester
You could go right to the definition

$
D'(-1) = \lim_{x \to -1} \frac{D(x) - D(-1)}{x+1} = \lim_{x \to -1} \frac{(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)}{x+1}
$

$= \lim_{x \to -1} (x+2)^2(x+3)^3(x+4)^4(x+5) = 1^2 \cdot 2^3 \cdot 3^4 \cdot 4$ as Skeeter said.
• Dec 20th 2009, 12:18 PM
oblixps
I'm curious, why didn't logarithmic differentiation work for this problem?
• Dec 20th 2009, 12:20 PM
VonNemo19
Quote:

Originally Posted by oblixps
I'm curious, why didn't logarithmic differentiation work for this problem?

I was just about to say that.
• Dec 20th 2009, 12:53 PM
skeeter
Quote:

Originally Posted by oblixps
I'm curious, why didn't logarithmic differentiation work for this problem?

actually, it does ... take the limit as $x \to e^{i \pi}$ of the log expression posted by e^(i*pi)
• Dec 20th 2009, 04:18 PM
oblixps
but ln(1+e^(i*pi)) = ln(0), doesn't the limit not exist?
• Dec 20th 2009, 05:21 PM
oblixps
Quote:

Originally Posted by e^(i*pi)
Use log differentiation

$ln[D(x)] = ln[(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]$ $= ln(x+1) + 2ln(x+2) + 3ln(x+3) + 4ln(x+4) + ln(x+5)$

Recall that due the chain rule: $\frac{d}{dx}ln[f(x)] = \frac{f'(x)}{f(x)}$

Actually, looking at it I don't think it is differentiable at x=-1 because of the 1/(x+1) term

after differentiating both sides and multiplying both sides by $(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)$, you'll get y'= $[\frac{1}{x+1} + \frac{2}{x+2}+\frac{3}{x+3}+\frac{4}{x+4}+\frac{1} {x+5}]$ $[(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]$

then distribute and plug in x = -1 and you'll see that all the terms will cancel out and become zero except for the first term, since the (x+1)s cancel out. when you plug in x = -1 to what you have left, you'll get $(1^2)(2^3)(3^4)(4)$
• Dec 21st 2009, 08:25 AM
dapore
You the most beautiful tribute (Clapping)
• Dec 21st 2009, 09:16 AM
Krahl
Quote:

Originally Posted by oblixps
after differentiating both sides and multiplying both sides by $(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)$, you'll get y'= $[\frac{1}{x+1} + \frac{2}{x+2}+\frac{3}{x+3}+\frac{4}{x+4}+\frac{1} {x+5}]$ $[(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]$

then distribute and plug in x = -1 and you'll see that all the terms will cancel out and become zero except for the first term, since the (x+1)s cancel out. when you plug in x = -1 to what you have left, you'll get $(1^2)(2^3)(3^4)(4)$

ye so whats this about there being a problem with 1/(1+x)? am i missing something(Worried)
• Dec 21st 2009, 11:32 AM
oblixps
e^(i*pi) said the derivative does not exist because of the 1/(x+1) term but in fact it just cancels out when you distribute so the derivative does exist. I was just correcting a small mistake.