Question

Find dy/dx. If y = (x-1)(x-2)/√x

Answer 1

Answer:

$\displaystyle \sf \frac{dy}{dx} = \frac{3x^2-3x-2}{2x^{3/2}}$

Explanation:

Given :

$\displaystyle \sf y =\frac{( x - 1 ) ( x - 2 )}{\sqrt{x}}$

$\displaystyle \sf y =\frac{( x - 2x-x+2 )}{\sqrt{x}}$

$\displaystyle \sf y =\frac{( x -3x+2 )}{\sqrt{x}}$

We are asked to find d y / d x

Using quotient rule here :

$\displaystyle \sf \frac{d}{dx} \left[\frac{f(x)}{g(x)} \right]=\frac{g(x).f'(x)-f(x).g'(x)}{\left(g(x)\right)^2}$

Diff. w.r.t. x :

We also know :

$\displaystyle \sf \frac{d}{dx} \sqrt{x}=\frac{1}{2\sqrt{x}}$

$\displaystyle \sf y = \frac{( x^2- 3 x + 2)}{\sqrt{x}}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{\sqrt{x}.( x^2- 3 x + 2)'-( x^2- 3 x + 2).(\sqrt{x})'}{(\sqrt{x})^2}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{\sqrt{x}.( x^2- 3 x + 2)'-( x^2- 3 x + 2).(\sqrt{x})'}{x}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{\sqrt{x}.( 2x- 3 + 0)-( x^2- 3 x + 2).(\sqrt{x})'}{x}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{\sqrt{x}.( 2x- 3)-[( x^2- 3 x + 2)/2\sqrt{x} \ ]}{x}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{(2\sqrt{x})\sqrt{x}.( 2x- 3)-( x^2- 3 x + 2)}{2x.\sqrt{x}}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{2x.( 2x- 3)-( x^2- 3 x + 2)}{2x^{3/2}}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{4x^2-6x-( x^2- 3 x + 2)}{2x^{3/2}}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{4x^2-6x- x^2+3 x-2}{2x^{3/2}}$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = \frac{3x^2-3x-2}{2x^{3/2}}$

Hence we get required answer.

Answer 2

$\large \underline{ \red{ \boxed{ \bf \orange{Required \: Answer}}}}$