Question

find the derivative of √2x-3+√7-3x

Find the derivative ​

Answer 1

Answer ,

Given equation,

$\longrightarrow \tt \: y = \sqrt{2x - 3} + \sqrt{7 - 3x}$

Since the it's a function in function

• First , we differentiate the main function and then the internal function

Differentiating on both the sides ,

$\longrightarrow \rm \: \frac{dy}{dx} = \frac{d}{dx} { \bigg\{(2x - 3)}^{ \frac{1}{2} } + {(7 - 3x)}^{ \frac{1}{2} } \bigg\} \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \rm \: \frac{dy}{dx} = \frac{d}{dx} \: {(2x - 3)}^{ \frac{1}{2} } + \frac{d}{dx} {(7 - 3x)}^{ \frac{1}{2} } \: \: \: \: \: \: \: \: \: \: \: \\ \\ \longrightarrow \rm \: \frac{dy}{dx} = \frac{1}{ \cancel2 \sqrt{2x - 3} } \times \cancel2 + \frac{1}{2 \sqrt{7 - 3x} } \times ( - 3) \\ \\ \longrightarrow { \underline{\rm{ \red{ \underline{ \: \frac{dy}{dx} = \frac{1}{ \sqrt{2x - 3} } - \frac{ - 3}{2 \sqrt{7 - 3x} } }}}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 2

Questions:-

find the derivative of √2x-3+√7-3x ?

Answer

We know

$\tt{\dfrac{d}{dx}( \sqrt{x} )} = \dfrac{d}{dx} ( {x}^{ \frac{1}{2} } ) = \frac{1}{2} {(x)}^{1 - \frac{1}{2} }$

$\tt = \dfrac{1}{2} {(x)}^{ - \frac{1}{2} } = \dfrac{1}{2 \sqrt{x} }$

So, using it as a identity of derivative...

Let's come to the question...

$= \tt \frac{d}{dx}( \sqrt{2x - 3} + \sqrt{7 - 3x} )$

$\tt = \frac{1}{2 \sqrt{2x - 3}}.\frac{d}{dx}(2x - 3) + \frac{1}{2 \sqrt{7 - 3x}}.\frac{d}{dx}(7 - 3x)$

$\tt= \frac{1}{ \cancel2 \sqrt{2x - 3} } . \cancel2 + \frac{1}{2 \sqrt{7 - 3x} } . - 3$

$\underline{\underline{\boxed{\tt= \frac{1}{ \sqrt{2x - 3} } - \frac{3}{2 \sqrt{7 - 3x} } }}}$

Some Derivative Formulae:-

$\tt \: \dfrac{d}{dx} ( {x)}^{n} = n {x}^{n - 1}$

$\tt\dfrac{d}{dx}( logx) = \dfrac{1}{x}$

$\tt \: \dfrac{d}{dx} (c) = 0$

$\tt\dfrac{d}{dx} (x) = 1$

$\tt \dfrac{d}{dx} (u \: \pm \: v) = \dfrac{du}{dx} \pm \: \dfrac{dv}{dx}$