   Question Answers

# Differentiate the following with respect to ${\text{x}}$$y = {\left( {{x^2} + 2} \right)^2}.\sin \,\,x$  Hint: Differentiation of function means to compute the derivative of that function. A derivative is the rate at which output changes with respect to an input. Use product rule $\left( {f\left( x \right) \times g\left( x \right)} \right)$ to differentiate the given value with respect to${\text{x}}$.

Complete step by step solution:
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
$\Rightarrow \left( {{x^2} + 2} \right)\,\,\left( {4x\sin x + \left( {{x^2} + 2} \right)\cos x} \right)$
$\dfrac{{dy}}{{dx}} = \left( {{x^2} + 2} \right)\,\,\left( {4x\sin x + \left( {{x^2} + 2} \right)\cos x} \right)$

Note: Students should follow product rule $\left[ {f\left( x \right)g\left( x \right)} \right]$when they differentiate this value with respect to ${\text{x}}$ then
$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right]$

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