# How do you find the derivative of y=[(2-3x^2)^(1/2)](5x+2)?

Jun 30, 2017

$\frac{- 15 {x}^{2} - 15 x + 4}{\sqrt{2 - 3 {x}^{2}}}$

#### Explanation:

The derivative of a product $\textcolor{red}{f} \left(x\right) \cdot \textcolor{g r e e n}{g} \left(x\right)$ is:

$f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

Let's assume that

$\textcolor{red}{f \left(x\right) = {\left(2 - 3 {x}^{2}\right)}^{\frac{1}{2}}}$

and

$\textcolor{g r e e n}{g \left(x\right) = 5 x + 2}$

Since

$f \left(x\right) = {\left({f}_{1} \left(x\right)\right)}^{a} \to f ' \left(x\right) = \textcolor{red}{a {\left({f}_{1} \left(x\right)\right)}^{a - 1}} \cdot \textcolor{b l u e}{{f}_{1} ' \left(x\right)}$

the derivative finally is:

$\textcolor{red}{\frac{1}{\cancel{2}} {\left(2 - 3 {x}^{2}\right)}^{\frac{1}{2} - 1} \cdot \textcolor{b l u e}{\left(- 3 \cdot \cancel{2} x\right)}} \cdot \textcolor{g r e e n}{\left(5 x + 2\right)} + \textcolor{red}{{\left(2 - 3 {x}^{2}\right)}^{\frac{1}{2}}} \cdot \textcolor{g r e e n}{5}$

$= - 3 \cdot {\left(2 - 3 {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(5 x + 2\right) + 5 \cdot {\left(2 - 3 {x}^{2}\right)}^{\frac{1}{2}}$

or

$- \frac{3 \left(5 x + 2\right)}{\sqrt{2 - 3 {x}^{2}}} + 5 \sqrt{2 - 3 {x}^{2}}$

$= \frac{- 3 \left(5 x + 2\right) + 5 \left(2 - 3 {x}^{2}\right)}{\sqrt{2 - 3 {x}^{2}}}$

$= \frac{- 15 x - 6 + 10 - 15 {x}^{2}}{\sqrt{2 - 3 {x}^{2}}}$

$= \frac{- 15 {x}^{2} - 15 x + 4}{\sqrt{2 - 3 {x}^{2}}}$