derivitives

**weezie23** · June 7th 2008, 01:54 PM

I have to find y'' for rad(t^3+1)

I got y' = 3t^2/(2rad(t^3+1)

i need help with y'' I cannot get to the answer

**o_O** · June 7th 2008, 02:03 PM

$y = \sqrt{t^{3} + 1}$

You calculated y' correctly. To calculate y'', you can use the quotient rule:
$y' = \frac{3}{2} \cdot \frac{t^{2}}{(t^{3}+1)^{\frac{1}{2}}}$
$y'' = \frac{3}{2} \cdot \frac{d}{dt}\left( \frac{t^{2}}{(t^{3}+1)^{\frac{1}{2}}}\right) = \frac{3}{2} \cdot \underbrace{\frac{\left(t^2\right)'(t^{3}+1)^{\fra c{1}{2}} \: \: - \: \: t^{2}\left[\left(t^3 + 1\right)^{\frac{1}{2}}\right]'}{\left[\left(t^3+1\right)^{\frac{1}{2}}\right]^{2}}}_{\text{Quotient rule}}$

In case you forgot: $\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{\left[g(x)\right]^2}$

**Mathstud28** · June 7th 2008, 02:06 PM

Originally Posted by weezie23

I have to find y'' for rad(t^3+1)

I got y' = 3t^2/(2rad(t^3+1)

i need help with y'' I cannot get to the answer

rad? If you mean

$\frac{d}{dx}\bigg[\sqrt{x^3+1}\bigg]$

Then let $y=\sqrt{x^3+1}\Rightarrow{y^2=x^3+1}$

Differentiating we get

$2y\cdot{y'}=3x^2$

Or $y'=\frac{3x^2}{2y}$

Remembering our original equation is y we get

$y'=\frac{3x^2}{2\sqrt{x^3+1}}$

So now we take the second derivative first rewriting this as

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

So $y''=\frac{3}{2}\bigg[2x(x^3+1)^{\frac{-1}{2}}+\frac{-x^2}{2}(x^3+1)^{\frac{-3}{2}}\bigg]$

I think you can go from there

**weezie23** · June 7th 2008, 02:07 PM

I included the 2/3 while calculating q. rule.

Thanks!

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