Question

find dy/dx if y=(x+1)/√x ,x=1/4​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow y'=-3$

Step-by-step explanation:

Given :

$\displaystyle \sf y=\frac{\left( x+1 \right)}{\sqrt{x}}$

We have to find d y / d x at x = 1 / 4 :

Using quotient rule :

= > d / d x [ f ( x ) / g ( x ) ] = [ g ( x ) . f' ( x ) - f ( x ) . g' ( x ) ] / [ g ( x ) ]²

$\displaystyle \sf y'=\frac{\sqrt{x}\left( x+1 \right)'-\left(x+1\right)\left(\sqrt{x}\right)'}{\left[\sqrt{x}\right]^2}$

$\displaystyle \sf y'=\frac{\sqrt{x}\left( x+1 \right)'-\left(x+1\right)\left(\sqrt{x}\right)'}{x}$

$\displaystyle \sf y'=\frac{\sqrt{x}\left( 1+0 \right)-\left(x+1\right)\left(\sqrt{x}\right)'}{x}$

$\displaystyle \sf y'=\frac{\sqrt{x}-\left(x+1\right)\left(\sqrt{x}\right)'}{x}$

$\displaystyle \sf y'=\frac{\sqrt{x}-\left(x+1\right)\left(\dfrac{1}{2\sqrt{x}}\right)}{x}$

$\displaystyle \sf y'=\frac{2x-\left(x+1\right)}{2x.\sqrt{x}}$

$\displaystyle \sf y'=\frac{x-1}{2x.\sqrt{x}}$

Now at x = 1 / 4 :

$\displaystyle \sf y'=\frac{\dfrac{1}{4} -1}{2.\dfrac{1}{4} .\sqrt{\dfrac{1}{4} }}$

$\displaystyle \sf y'=\frac{\dfrac{-3}{4}}{2.\dfrac{1}{4} .\dfrac{1}{2} }$

$\displaystyle \sf y'=\frac{\dfrac{-3}{4}}{\dfrac{1}{4} }$

$\displaystyle \sf \longrightarrow y'=-3$

Hence we get required answer!