Question

Q1.find the derivative of given functions with respect to independent variable :
Y= (x-1) (x²+x+1)
​

Answer 1

Step-by-step explanation:

⟹y=(x−1)(x² +x+1)

On differentiating both sides w.r.t x :-

★

★ By using Product Rule :

★ By using Product Rule :

$</p><p>\begin{gathered}:\implies \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = u\dfrac{dv}{dx}+ v\dfrac{du}{dx} \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = (x - 1)\dfrac{d}{dx}( {x}^{2} + x + 1)+ ( {x}^{2} + x + 1) \dfrac{d}{dx}(x - 1) \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = (x - 1)\dfrac{d}{dx}( 2{x}^{2 - 1} + {x}^{1 - 1} + 0)+ ( {x}^{2} + x + 1) \dfrac{d}{dx}( {x}^{1 - 1} - 0) \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = (x - 1)( 2 {x}^{1} + 1)+ ( {x}^{2} + x + 1)(1) \\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = {2x}^{2} + x - 2x - 1+ {x}^{2} + x + 1\\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = {2x}^{2} -x - 1+ {x}^{2} + x + 1\\ \\ \end{gathered} </p><p>$

$\begin{gathered}:\implies \underline{ \boxed{ \sf \dfrac{dy}{dx} = \dfrac{d}{dx}(u.v) = {3x}^{2} }}\\ \end{gathered} </p><p>$

Answer 2

Answer:

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