# How do you find the derivative of sqrt(5-3x)?

Nov 5, 2016

$= \frac{- 3}{2 \sqrt{5 - 3 x}}$

#### Explanation:

To find the derivative of the expression is by applying chain rule
Let $v \left(x\right) = 5 - 3 x \mathmr{and} u \left(x\right) = \sqrt{x}$

Then $u \circ v \left(x\right) = u \left(v \left(x\right)\right) = \sqrt{5 - 3 x}$

Then color(blue)((sqrt(5-3x))'=(u@v(x))'=u'(v(x))xxv'(x)

Computing $u ' \left(v \left(x\right)\right) \mathmr{and} v ' \left(x\right)$

$u ' \left(x\right) = \frac{1}{2 \sqrt{x}}$
Then

color(blue)( u'(v(x))=1/(2sqrt(5-3x))

color(blue)(v'(x)=-3

color(blue)((sqrt(5-3x))'=(u@v(x))'=u'(v(x))xxv'(x)

$\left(\sqrt{5 - 3 x}\right) ' = \frac{1}{2 \sqrt{5 - 3 x}} \times - 3$

Therefore,
$\left(\sqrt{5 - 3 x}\right) ' = \frac{- 3}{2 \sqrt{5 - 3 x}}$

Nov 5, 2016

 d/dx sqrt(5-3x) = -3/(2sqrt(5-3x)

#### Explanation:

$f \left(x\right) = \sqrt{5 - 3 x}$
We know,
$\frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) = \frac{1}{2 \sqrt{x}}$
and also,
$\frac{d}{\mathrm{dx}} \left(a - b x\right) = - b$

hence, using the chain rule, we differentiate $f \left(x\right)$ to get
 d/dx sqrt(5-3x) = -3/(2sqrt(5-3x)

Nov 5, 2016

The answer is $= - \frac{3}{2 \sqrt{5 - 3 x}}$

#### Explanation:

For this, we use $\left(\sqrt{u}\right) ' = \frac{1}{2 \sqrt{u}}$ and the chain rule

So, $\left(\sqrt{5 - 3 x}\right) ' = \frac{1}{2 \sqrt{5 - 3 x}} \cdot \left(- 3 x\right) '$

$= - \frac{3}{2 \sqrt{5 - 3 x}}$

Nov 5, 2016

$- \frac{3}{2 \sqrt{5 - 3 x}}$

#### Explanation:

Express $y = \sqrt{5 - 3 x} = {\left(5 - 3 x\right)}^{\frac{1}{2}}$

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{2}{2}} |}}} \to \left(A\right)$

let $u = 5 - 3 x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = - 3$

and $y = {u}^{\frac{1}{2}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2} {u}^{- \frac{1}{2}}$

substitute these values into (A) changing u back into terms of x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {u}^{- \frac{1}{2}} \times \left(- 3\right) = - \frac{3}{2 {u}^{\frac{1}{2}}} = - \frac{3}{2 \sqrt{5 - 3 x}}$