# How do you differentiate (x^2+3)sqrt(x+2)?

Mar 13, 2018

$\frac{d}{\mathrm{dx}} \left(\left({x}^{2} + 3\right) \sqrt{x + 2}\right) = \frac{5 {x}^{2} + 8 x + 3}{2 \sqrt{x + 2}}$

#### Explanation:

Using the product rule:

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

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

Now simplifying:

$\frac{d}{\mathrm{dx}} \left(\left({x}^{2} + 3\right) \sqrt{x + 2}\right) = \frac{4 x \left(x + 2\right) + \left({x}^{2} + 3\right)}{2 \sqrt{x + 2}}$

$\frac{d}{\mathrm{dx}} \left(\left({x}^{2} + 3\right) \sqrt{x + 2}\right) = \frac{4 {x}^{2} + 8 x + {x}^{2} + 3}{2 \sqrt{x + 2}}$

$\frac{d}{\mathrm{dx}} \left(\left({x}^{2} + 3\right) \sqrt{x + 2}\right) = \frac{5 {x}^{2} + 8 x + 3}{2 \sqrt{x + 2}}$