What is the second derivative of y=x*sqrt(16-x^2)?

Aug 22, 2015

${y}^{' '} = \frac{2 \cdot x \left({x}^{2} - 24\right)}{\left(16 - {x}^{2}\right) \cdot \sqrt{16 - {x}^{2}}}$

Explanation:

Start by calculating the first derivative of your function $y = x \cdot \sqrt{16 - {x}^{2}}$ by using the product rule.

This will get you

$\frac{d}{\mathrm{dx}} \left(y\right) = \left[\frac{d}{\mathrm{dx}} \left(x\right)\right] \cdot \sqrt{16 - {x}^{2}} + x \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{16 - {x}^{2}}\right)$

You can differentiate $\frac{d}{\mathrm{dx}} \left(\sqrt{16 - {x}^{2}}\right)$ by using the chain rule for $\sqrt{u}$, with $u = 16 - {x}^{2}$.

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{d}{\mathrm{du}} \sqrt{u} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{u}} \cdot \frac{d}{\mathrm{dx}} \left(16 - {x}^{2}\right)$

$\frac{d}{\mathrm{dx}} \left(\sqrt{16 - {x}^{2}}\right) = \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \frac{1}{\sqrt{16 - {x}^{2}}} \cdot \left(- \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x\right)$

$\frac{d}{\mathrm{dx}} \left(\sqrt{1 - {x}^{2}}\right) = - \frac{x}{\sqrt{16 - {x}^{2}}}$

Plug this back into your calculation of ${y}^{'}$.

${y}^{'} = 1 \cdot \sqrt{16 - {x}^{2}} + x \cdot \left(- \frac{x}{\sqrt{16 - {x}^{2}}}\right)$

${y}^{'} = \frac{1}{\sqrt{16 - {x}^{2}}} \cdot \left(16 - {x}^{2} - {x}^{2}\right)$

${y}^{'} = \frac{2 \left(8 - {x}^{2}\right)}{\sqrt{16 - {x}^{2}}}$

To find ${y}^{' '}$ you need to calculate $\frac{d}{\mathrm{dx}} \left({y}^{'}\right)$ by using the quotient rule

$\frac{d}{\mathrm{dx}} \left({y}^{'}\right) = 2 \cdot \frac{\left[\frac{d}{\mathrm{dx}} \left(8 - {x}^{2}\right)\right] \cdot \sqrt{16 - {x}^{2}} - \left(8 - {x}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{16 - {x}^{2}}\right)}{\sqrt{16 - {x}^{2}}} ^ 2$

${y}^{' '} = 2 \cdot \frac{- 2 x \cdot \sqrt{16 - {x}^{2}} - \left(8 - {x}^{2}\right) \cdot \left(- \frac{x}{\sqrt{16 - {x}^{2}}}\right)}{16 - {x}^{2}}$

${y}^{' '} = 2 \cdot \frac{\frac{1}{\sqrt{16 - {x}^{2}}} \cdot \left[- 2 x \cdot \left(16 - {x}^{2}\right) + x \cdot \left(8 - {x}^{2}\right)\right]}{16 - {x}^{2}}$

${y}^{' '} = \frac{2}{\sqrt{16 - {x}^{2}} \cdot \left(16 - {x}^{2}\right)} \cdot \left(- 32 x + 2 {x}^{3} + 8 x - {x}^{3}\right)$

Finally, you have

${y}^{' '} = \textcolor{g r e e n}{\frac{2 \cdot x \left({x}^{2} - 24\right)}{\left(16 - {x}^{2}\right) \cdot \sqrt{16 - {x}^{2}}}}$