# What is the derivative of the function ? y = x √ 2x + 3

done it for you

#### Explanation:

Here I have used the uv rule for differentiation

Mar 8, 2018

solution of $y = x \sqrt{2 x + 3}$ instead of $y = x \sqrt{2} x + 3$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 \left(x + 1\right)}{\sqrt{2 x + 3}}$
HINT:
color(red)((1)y=u*v=>(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)

#### Explanation:

$y = x \cdot \sqrt{2 x + 3} = x \cdot {\left(2 x + 3\right)}^{\frac{1}{2}}$$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{d}{\mathrm{dx}} \left({\left(2 x + 3\right)}^{\frac{1}{2}}\right) + {\left(2 x + 3\right)}^{\frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(x\right)$
$= x \cdot \frac{1}{\cancel{2}} \cdot {\left(2 x + 3\right)}^{\frac{1}{2} - 1} \cdot \cancel{2} + {\left(2 x + 3\right)}^{\frac{1}{2}}$
=x(2x+3)^(-1/2)+(2x+3)^(1/2
$= \frac{x}{\sqrt{2 x + 3}} + \sqrt{2 x + 3}$
$= \frac{x + 2 x + 3}{\sqrt{2 x + 3}} = \frac{3 x + 3}{\sqrt{2 x + 3}} = \frac{3 \left(x + 1\right)}{\sqrt{2 x + 3}}$
We can multiply $x$ with $\sqrt{2 x + 3}$
$y = x {\left(2 x + 3\right)}^{\frac{1}{2}} = {\left({x}^{2}\right)}^{\frac{1}{2}} {\left(2 x + 3\right)}^{\frac{1}{2}} = {\left[{x}^{2} \left(2 x + 3\right)\right]}^{\frac{1}{2}}$$\implies y = {\left(2 {x}^{3} + 3 {x}^{2}\right)}^{\frac{1}{2}} \implies {y}^{'} = \frac{1}{2} {\left(2 {x}^{3} + 3 {x}^{2}\right)}^{1 - \frac{1}{2}} \left(6 {x}^{2} + 6 x\right)$
${y}^{'} = \frac{6 x}{2} {\left(2 {x}^{3} + 3 {x}^{2}\right)}^{- \frac{1}{2}} \left(x + 1\right) = \frac{3 x \left(x + 1\right)}{\sqrt{2 {x}^{3} + 3 {x}^{2}}} = \frac{3 x \left(x + 1\right)}{\sqrt{{x}^{2} \left(2 x + 3\right)}} = \frac{3 x \left(x + 1\right)}{x \sqrt{\left(2 x + 3\right)}} = \frac{3 \left(x + 1\right)}{\sqrt{2 x + 3}}$
Note:: $\to x \cdot \sqrt{2 x + 3} \ne x + \sqrt{2 x + 3}$