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Question

Differentiate the function $$ y = \dfrac{(x + 2)(3x -1)}{(2x + 5)} $$ with respect to $$ x $$ 


Solution

$$ y = \dfrac{(x + 2)(3x -1)}{(2x + 5)} $$ 
$$ \dfrac{dy}{dx} = \dfrac{d}{dx} \dfrac{(x + 2)(3x - 1)}{(2x + 5)} $$ 
$$ = \dfrac{(2x + 5) \dfrac{d}{dx} (x + 2)(3x - 1) - (x + 2) (3x - 1) \dfrac{d}{dx} (2x + 5)}{(2x + 5)^2} $$ 
$$ = \dfrac{(2x + 5) \left  [ (x + 2) \dfrac{d}{dx} (3x - 1) + (3x - 1) \dfrac{d}{dx} (x + 2)  \right  ] - (x + 2)(3x - 1)[2 + 0]}{(2x + 5)^2} $$ 
$$ = \dfrac{(2x + 5) [(x + 2) \times 3 + (3x - 1) \times 1] - 2[3x^2 + 6x - x - 2]}{(2x + 5)^2} $$  
$$ = \dfrac{(2x + 5)[3x + 6 + 3x - 1] - 6x^2 - 12x + 2x + 4}{(2x + 5)^2} $$ 
$$ = \dfrac{12x^2 + 30x + 10x + 25 - 6x^2 - 10x + 4}{(2x + 5)^2} $$ 
$$ = \dfrac{6x^2 + 30x + 29}{(2x + 5)^2} $$

Mathematics

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