Question

y=3x2+2x+6 then dy/dx=​

Answer 1

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GIVEN :

$\mapsto \sf \: y = 3 {x}^{2} + 2x + 6$

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SOLUTION :

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: y \: = 3 {x}^{2} + 2x + 6$

$\implies \sf \: \frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2} + 2x + 6)$

$\implies \sf \: \frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2} ) + \frac{d}{dx} (2x) + \frac{d}{dx} (6)$

$\implies \sf \: \frac{dy}{dx} = 6x + 2+ 0$

$\: \: \: \: \: \: \: \: \: \therefore { \underline{\boxed{\bold{\sf \: \frac{dy}{dx} = 6x + 2} }}}$

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REQUIRED CONCEPTS :

$\sf \mapsto \: \frac{d}{dx} ( {x}^{n} )= n \: {a}^{n - 1}$

$\mapsto \: \sf \frac{d}{dx} (c) = 0$

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CONCEPT BOOSTER :

$\sf 1) \: \: \frac{d}{dx} ( \sin \: x) = \cos \: x$

$\sf 2) \frac{d}{dx} ( \cos \: x) = - \sin \: x$

$\sf </u><u> 3) \: \: \frac{d}{dx} ( \tan \: x) = { \sec}^{2} x$

$\sf 4) \: \: \frac{d}{dx}( \cot \: x) = - { \cosec}^{2} x$

$\sf 5) \: \: \frac{d}{dx} ( \sec \: x) = \sec \: x \: \tan \: x$

$\sf 6) \: \: \frac{ d}{dx} \cosec \: x = - \cosec \: x \cot \: x$

$\sf</u><u> 7) \: \: \frac{d}{dx} ( {e}^{x} ) = {e}^{x}$

$\sf 8) \: \: \: \: \frac{d}{dx} ( log \: x) = \frac{1}{x} \: \: \: [ \: x \not = 0]$

$\sf 9) \: \: \frac{d}{dx} (u±v±w ±...) = \frac{d}{dx} u± \frac{d}{dx} v± \frac{d}{dx} w±...$

$\sf 10) \: \: \frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

$\sf 11) \: \: \frac{d}{dx} ( \frac{u}{v} ) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {v}^{2} }$

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