Question

find dy / dx if y = (x-1) (x-2)/√x
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Answer 1

$\sf\large\red{\underline{\underline{Question:}}}$

• Find dy / dx if y = (x-1) (x-2)/√x

$\sf\large\red{\underline{\underline{Solution:}}}$

We have ,

$\sf \longmapsto \blue{y= \dfrac{ (x-1)(x-2)}{\sqrt x}}$

$\dashrightarrow \sf y = \dfrac{ x^2-2x-x+2}{\sqrt x}$  

$\dashrightarrow \sf y = \dfrac{ x^2-3x+2}{\sqrt x}$  

$\large\purple{\underbrace{\mathfrak{Diferentiating\;both\;sides\;with\;respect\;to\; \sf x : }}}$

$\\\dfrac{dy}{dx}=\dfrac{\sqrt{x}\dfrac{d}{dx}(x^2-3x+2)-(x^2-3x+2)\dfrac{d}{dx}\sqrt x}{\sqrt x ^2}$

$\dfrac{dy}{dx}=\dfrac{\sqrt x (2x-3)-(x^2-3x+2)\dfrac{1}{2}.x^{\frac{-1}{2}}}{x}$

$\dfrac{dy}{dx}=\dfrac{\sqrt x (2x-3)-(x^2-3x+2)\dfrac{1}{2\sqrt x}}{x}$

$\dfrac{dy}{dx}=\dfrac{\dfrac{2\sqrt x.\sqrt x (2x-3)-(x^2-3x+2)}{2\sqrt x}}{x}$

$\dfrac{dy}{dx}=\dfrac{\dfrac{2x(2x-3)-(x^2-3x+2)}{2\sqrt x}}{x}$

$\dfrac{dy}{dx}=\dfrac{2x(2x-3)-(x^2-3x+2)}{2x\sqrt x}$

$\dfrac{dy}{dx}=\dfrac{4x^2-6x-x^2+3x-2}{2x\sqrt x}$

$\dfrac{dy}{dx}= \dfrac{3x^2-3x-2}{2x\sqrt x}$

$\\\succ\large\purple{\underbrace{\mathfrak{Rationalising:}}}$

$\longmapsto \dfrac{3x^2-3x-2}{2x\sqrt x}\times\dfrac{\sqrt x}{\sqrt x}$

$\dashrightarrow \dfrac{\sqrt x(3x^2-3x-2)}{2x^2}$

$\sf\large\red{\underline{\underline{Therefore,}}}$

• $\dfrac{dy}{dx}= \dfrac{\sqrt x(3x^2-3x-2)}{2x^2}$

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$\sf\large\red{\underline{\underline{Formulas\: Used:}}}$

• (U + V)' = U' + V'
• (UV)'= U'V + V'U
• $\sf\left(\dfrac{U}{V}\right)'= \dfrac{V(U)'-U(V)'}{V^2}$
• $\dfrac{d}{dx}\sf x^n= n.x^{n-1}$