Nasty derivative

**ffezz** · March 18th 2008, 09:01 PM

Once again. After missing a week of school and a test, its not fun to catch up, especially in calculus.

Heres the question.

Find the slope of the tangent to $h(x)=2x(x+1)^3(x^2+2x+1)^2$ at
$x=-2$
Explain how to find the equation of the normal at x= -2.

Ok so if anyone solves this, can they please do a step by step break down because im about to eat my book.

**iknowone** · March 18th 2008, 09:23 PM

$<br /> h(x)=2x(x+1)^3(x^2+2x+1)^2<br />$

$g(x)=2x(x+1)^3 = q(x)p(x)$
where $q(x) = 2x$
and $p(x) =(x+1)^3$

$f(x)=(x^2+2x+1)^2$

$f'(x) = 2(x^2+2x+1)(2x+2)$

$g'(x) = q(x)p'(x) + q'(x)p(x) = 2x(3(x+1)^2) + 2(x+1)^3$
_________________________________________________

$h(x) = g(x)f(x) \Rightarrow h'(x) = g(x)f'(x) + g'(x)f(x)$

$h(x) = 2x(x+1)^3 2(x^2+2x+1)(2x+2) + (2x(3(x+1)^2) + 2(x+1)^3)(x^2+2x+1)^2$

simplify and plug in x=-2!

oh and check my work, this thing's a mess!

**ffezz** · March 18th 2008, 09:46 PM

eeek. i just worked through it and it gave me -24. the answer in the book says -30. Il double check my math

**TheEmptySet** · March 18th 2008, 09:55 PM

let

$h(x)=2x(x+1)^3(x^2+2x+1)^2<br />$

so $h(-2)=1(-2)(-1)(1)^2=4$

talking the log of both sides we get

$ln(h(x))=ln(2x)+3ln(x+1)+2ln(x^2+2x+1)$ take the derivative

$\frac{1}{h}\frac{dh}{dx}=\frac{1}{x}+\frac{3}{x+1} +\frac{2(2x+2)}{x^2+2x+1}$ solving for the derivative...

$\frac{dh}{dx}=h(x)[\frac{1}{x}+\frac{3}{x+1}+\frac{2(2x+2)}{x^2+2x+1}]$ eval at x=-2

$\frac{dh}{dx}|_{x=-2}=h(-2)[ \frac{1}{-2}+\frac{3}{-2+1}+\frac{2(2(-2)+2)}{(-2)^2+2(-2)+1}]$

$\frac{dh}{dx}|_{x=-2}=4[\frac{-1}{2}-3-4]=-30$

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