First and Second Derivatives

**midsummer** · October 9th 2008, 10:45 AM

Find the first and second derivatives of the function y=4x/ square root (X+1)

Thanks!

**Moo** · October 9th 2008, 10:50 AM

Hello,

Originally Posted by midsummer

Find the first and second derivatives of the function y=4x/ square root (X+1)

Thanks!

$f(x)=\frac{4x}{\sqrt{x+1}}=4x \cdot (x+1)^{-1/2}$ (remember that sqrt(x)=x^(1/2))

Note that the constant is not important, because $[af(x)]'=af'(x)$ for any constant.

Apply product rule :
$f'(x)=4 (x+1)^{-1/2}+4x \cdot \tfrac{-1}{2} \cdot (x+1)^{-1/2-1}$ $=4(x+1)^{-1/2}-2x (x+1)^{-3/2}=2(x+1)^{-3/2} \big[2(x+1)-x\big]$

$f'(x)=2(x+1)^{-3/2} \big[x+2\big]$

Apply product rule again to get $f''(x)$

**icemanfan** · October 9th 2008, 10:52 AM

You will have to use the quotient rule for this question.

The quotient rule is $\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$ .

We have: $y = \frac{4x}{\sqrt{x+1}}$

Therefore, $y' = \frac{\sqrt{x+1} \cdot 4 - 4x \cdot (1/2)(x+1)^{-1/2}}{x + 1} = \frac{4(x + 1) - 2x}{(x + 1)^{3/2}} = \frac{2x + 4}{(x + 1)^{3/2}}$ .

This is the first derivative.

To obtain the second derivative, you must use the quotient rule to determine the derivative of the first derivative.

**midsummer** · October 9th 2008, 12:06 PM

thanks...and could you demonstrate how to find the second derivative? I understand the concept and I am trying to simplify (it's been a while since I have done this stuff and need lots of practice). again I really appreciate the help!

Mariam

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