2nd Deriv to Pt Inflect (number 2)

**akuczma86** · April 13th 2010, 10:47 AM

3x/(x+8).

I need to use the 2nd derivative test to find the point(s) of inflection. I get confused about how to do derivatives of examples like this.

**harish21** · April 13th 2010, 10:47 AM

Originally Posted by akuczma86

3x/(x+8).

I need to use the 2nd derivative test to find the point(s) of inflection. I get confused about how to do derivatives of examples like this.

Quotient rule

If $f(x) = \frac{x}{y}$

then, $f'(x) = \frac{x'y-y'x}{y^2}$

This should definitely be in your books/notes

**akuczma86** · April 13th 2010, 10:50 AM

So the first derivative would be -3x/(x+8)^2?

**harish21** · April 13th 2010, 10:56 AM

Originally Posted by akuczma86

So the first derivative would be -3x/(x+8)^2?

NO.

Its

$\frac {([\text{derivative of 3x}]\times (x+8)) - (3x \times [\text{derivative of} (x+8)])}{(x+8)^2}$

**earboth** · April 13th 2010, 10:57 AM

Originally Posted by akuczma86

So the first derivative would be -3x/(x+8)^2? <<<<<< no

$f(x)=\frac{3x}{x+8}$

$f'(x)=\frac{(x+8) \cdot 3 -3x \cdot 1}{(x+8)^2}$

Expand the bracket in the numerator, collect like terms.

**akuczma86** · April 13th 2010, 11:17 AM

$f'(x)= 24/ (x^2 + 16x + 64$

$f''(x)= (-24 -3x)/(x^4 + 2x^3 + 19x^2 + 48x + 36)$

Am I right here?

**running-gag** · April 13th 2010, 11:26 AM

It is OK for f'(x)

BTW an easiest way in this case is $f(x) = \frac{3x}{x+8} = 3 - \frac{24}{x+8}$ . The derivative of f is the same as the derivative of $-\frac{24}{x+8}$ which is $\frac{24}{(x+8)^2}$

The derivative of $u^n(x)$ is $n u'(x) u^{n-1}(x)$

Use this formula with u(x) = x+8 and n=-2 to get the derivative of $\frac{1}{(x+8)^2}$ then multiply by 24

**earboth** · April 13th 2010, 10:58 PM

Originally Posted by akuczma86

$f'(x)= 24/ (x^2 + 16x + 64$

$f''(x)= (-24 -3x)/(x^4 + 2x^3 + 19x^2 + 48x + 36)$

Am I right here? <<<<<< I can't imagine how you got this result

$f'(x)= \frac{24}{x^2 + 16x + 64}$ then

$f''(x)= \frac{(x^2+16x+64) \cdot 0 - 24(2x+16)}{ (x^2 + 16x + 64)^2} = \boxed{\frac{ - 48x-384}{ x^4 + 32·x^3 + 384·x^2 + 2048·x + 4096} }$

This result is correct but nearly useless. Compare with the method posted by running-gag:

$f'(x)= \frac{24}{x^2 + 16x + 64} = 24 (x+8)^{-2}$ then

$f'(x)= -48 (x+8)^{-3} \cdot 1 =\boxed{ -\frac{48}{(x+8)^3}}$

Both terms describe the same function!

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