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\[y = {\left( {{x^2} + 2} \right)^2}.\sin \,\,x\]

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Given, \[y = {\left( {{x^2} + 2} \right)^2}.\sin x\]

Differentiate it with respect to x

$ \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}{\left( {{x^2} + 2} \right)^2}.\sin x $

When we differentiate the first value $ {({x^2} + 2)^2} $ then next value is constant in same manner when we differentiate 2nd value $ (\sin x) $ then 1st value will consider as a constant term

$

\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {{x^2} + 2} \right)^2}.\sin x \\

= \sin x\dfrac{d}{{dx}}{({x^2} + 2)^2} \times {\left( {{x^2} + 2} \right)^2}\dfrac{d}{{dx}}\sin x \\

$

While, differentiate $ {\left( {{x^2} + 2} \right)^2} $ then, remove, power further, will some.

$

\dfrac{{dy}}{{dx}} = \sin x\left( {{x^2} + 2} \right)\dfrac{d}{{dx}}\left( {{x^2} + 2} \right) + {\left( {{x^2} + 2} \right)^2}\cos x \\

\dfrac{{dy}}{{dx}} = \sin x({x^2} + 2)\left[ {\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( 2 \right)} \right] + {\left( {{x^2} + 2} \right)^2}\cos x \\

\dfrac{{dy}}{{dx}} = \sin x\left( {{x^2} + 2} \right)\left[ {2x + 0} \right] + {\left( {{x^2} + 2} \right)^2}\cos x \\

$ ( $\because (2) $ is constant we cannot do differentiation)

\[\dfrac{{dy}}{{dx}} = 2\left( {{x^2} + 2} \right).\left( {2x + 0} \right).\sin x + {\left( {{x^2} + 2} \right)^2}.\cos x\]

\[ \Rightarrow \,4x\left( {{x^2} + 2} \right)\sin x + {\left( {{x^2} + 2} \right)^2}\cos x\]

Taking common $\left( {{x^2} + 2} \right)$on both sides, we get

\[ \Rightarrow \left( {{x^2} + 2} \right)\,\,\left( {4x\sin x + \left( {{x^2} + 2} \right)\cos x} \right)\]

$ \dfrac{{dy}}{{dx}} = \left( {{x^2} + 2} \right)\,\,\left( {4x\sin x + \left( {{x^2} + 2} \right)\cos x} \right) $

$ \dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] $

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