Question

differentiate if y=(x-1) (x-2)/rootx​

Answer 1

Step-by-step explanation:

y=(x-1)(x-2)/√x

dy/dx={√xd/dx(x-1)(x-2) - (x-1)(x-2)d/dx√x} ÷x

={√x×2 -(x^2-3x+2)} ÷2√x ×x

=2√x/2√x×x -(x^2-3x+2)/2√x×x

=1/x -(x-1)(x-2)/2√x×x

Answer 2

Therefore the differentiation of the given function y = ( x-1 ) ( x-2 ) / √x is 'dy/dx = 5√x/2 -1/√x + 1/$x^3^/^2$'.

Given:

Function: y = ( x-1 ) ( x-2 ) / √x

To Find:

The differentiation of the given function y = ( x-1 ) ( x-2 ) / √x

Solution:

The given question can be solved as shown below.

Given function: y = ( x-1 ) ( x-2 ) / √x

( x-1 ) ( x-2 ) = x² - 2x -x + 2 = x² - 2x + 2

Now y = x² - 2x + 2 / √x

Differentiating on both sides with respect to 'x',

⇒ dy/dx = d ( x² - 2x + 2 / √x ) / dx

The given function is in the form of ( u/v ) then its differentiation becomes

→ d ( u/v ) / dx= ( v du/dx - u dv/dx ) / v²

Where u = x² - 2x + 2; v = √x

du/dx = 2x + 2; dv/dx = -1/2√x

⇒ dy/dx = d ( x² - 2x + 2 / √x ) / dx = [ √x × d ( x² - 2x + 2 ) / dx - ( x² - 2x + 2 ) × d ( √x ) / dx ] / √x²

⇒ dy/dx = [ √x × ( 2x + 2 ) - ( x² - 2x + 2 ) × ( -1/2√x ) ] / √x²

⇒ dy/dx = [ 2$x^3^/^2$ + 2√x - (-$x^3^/^2$ + 2√x - 2/√x ) / 2 ] / x

⇒ dy/dx = 2√x + 2/√x + √x/2 - 1/√x + 1/$x^3^/^2$

⇒ dy/dx = 5√x/2 -1/√x + 1/$x^3^/^2$

Therefore the differentiation of the given function y = ( x-1 ) ( x-2 ) / √x is 'dy/dx = 5√x/2 -1/√x + 1/$x^3^/^2$'.

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