How do you find the derivative of y=(1/3) ((x^2)+2)^(3/2)?

Nov 27, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \sqrt{{x}^{2} + 2}$

Explanation:

Rewriting slightly:

$y = \frac{1}{3} {\left({x}^{2} + 2\right)}^{\frac{3}{2}}$

Notice that ${\left({x}^{2} + 2\right)}^{\frac{3}{2}}$ is a function in the form ${u}^{\frac{3}{2}}$. The power rule and the chain tells us that we differentiate this normally, in that the derivative of ${x}^{\frac{3}{2}}$ is $\frac{3}{2} {x}^{\frac{1}{2}}$, but when there's a function in the middle we will multiply by the derivative of the inner function. Thus, the derivative of ${u}^{\frac{3}{2}}$ is $\frac{3}{2} {u}^{\frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$.

So, we see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \left(\frac{3}{2} {\left({x}^{2} + 2\right)}^{\frac{1}{2}}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 2\right)$

The power rule tells us that the derivative of ${x}^{2} + 2$ is $2 x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} \left(\frac{3}{2}\right) \left(2 x\right) {\left({x}^{2} + 2\right)}^{\frac{1}{2}}$

Simplifying:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \sqrt{{x}^{2} + 2}$