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

Question

Evelyn Agard · Accepted Answer

#y^('') = (2 * x(x^2 - 24))/((16-x^2) * sqrt(16-x^2))#

Start by calculating the first derivative of your function #y = x * sqrt(16-x^2)# by using the product rule.

This will get you

#d/dx(y) = [d/dx(x)] * sqrt(16 - x^2) + x * d/dx(sqrt(16 - x^2))#

You can differentiate #d/dx(sqrt(16 -x^2))# by using the chain rule for #sqrt(u)#, with #u = 16 -x^2#.

#d/dx(sqrt(u)) = d/(du)sqrt(u) * d/dx(u)#

#d/dx(sqrt(u)) = 1/2 * 1/sqrt(u) * d/dx(16-x^2)#

#d/dx(sqrt(16-x^2)) = 1/color(red)(cancel(color(black)(2))) * 1/sqrt(16-x^2) * (-color(red)(cancel(color(black)(2)))x)#

#d/dx(sqrt(1-x^2)) = -x/sqrt(16-x^2)#

Plug this back into your calculation of #y^'#.

#y^' = 1 * sqrt(16-x^2) + x * (-x/sqrt(16-x^2))#

#y^' = 1/sqrt(16-x^2) * (16-x^2 - x^2)#

#y^' = (2(8-x^2))/sqrt(16-x^2)#

To find #y^('')# you need to calculate #d/dx(y^')# by using the quotient rule

#d/dx(y^') = 2 * ([d/dx(8-x^2)] * sqrt(16-x^2) - (8-x^2) * d/dx(sqrt(16-x^2)))/(sqrt(16-x^2))^2#

#y^('') = 2 * (-2x * sqrt(16-x^2) - (8-x^2) * (-x/sqrt(16-x^2)))/(16-x^2)#

#y^('') = 2 * (1/sqrt(16-x^2) * [-2x * (16-x^2) +x * (8-x^2)])/(16-x^2)#

#y^('') = 2/(sqrt(16-x^2) * (16-x^2)) * (-32x + 2x^3 + 8x - x^3)#

Finally, you have

#y^('') = color(green)((2 * x(x^2 - 24))/((16-x^2) * sqrt(16-x^2)))#

Ezra Adams · Answer

undefined To find the second derivative of $ y = x \sqrt{16 - x^2} $, we first need to find the first derivative, and then differentiate again.

The first derivative, $ y' $, is calculated using the product rule:
$$ y' = (x)' \cdot \sqrt{16 - x^2} + x \cdot (\sqrt{16 - x^2})' $$
$$ y' = 1 \cdot \sqrt{16 - x^2} + x \cdot \frac{1}{2\sqrt{16 - x^2}} \cdot (-2x) $$
$$ y' = \sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}} $$

Now, we'll differentiate $ y' $ to find the second derivative, $ y'' $:
$$ y'' = \left( \sqrt{16 - x^2} \right)' - \left( \frac{x^2}{\sqrt{16 - x^2}} \right)' $$
$$ y'' = \frac{1}{2\sqrt{16 - x^2}} \cdot (-2x) - \frac{\sqrt{16 - x^2} \cdot 2x}{(16 - x^2)^{3/2}} $$
$$ y'' = -\frac{x}{\sqrt{16 - x^2}} - \frac{2x^2}{(16 - x^2)^{3/2}} $$

So, the second derivative of $ y = x \sqrt{16 - x^2} $ is $ y'' = -\frac{x}{\sqrt{16 - x^2}} - \frac{2x^2}{(16 - x^2)^{3/2}} $.