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Question: Differentiate the following with respect to \[{\text{x}}\] \[y = {\left( {{x^2} + 2} \right)^2}.\s...

Differentiate the following with respect to x{\text{x}}
y=(x2+2)2.sin⁡  xy = {\left( {{x^2} + 2} \right)^2}.\sin \,\,x

Explanation

Solution

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 (f(x)×g(x))\left( {f\left( x \right) \times g\left( x \right)} \right) to differentiate the given value with respect tox{\text{x}}.

Complete step by step solution:
Given, y=(x2+2)2.sin⁡xy = {\left( {{x^2} + 2} \right)^2}.\sin x
Differentiate it with respect to x
ddxy=ddx(x2+2)2.sin⁡x\dfrac{d}{{dx}}y = \dfrac{d}{{dx}}{\left( {{x^2} + 2} \right)^2}.\sin x
When we differentiate the first value (x2+2)2{({x^2} + 2)^2} then next value is constant in same manner when we differentiate 2nd value (sin⁡x)(\sin x) then 1st value will consider as a constant term
dydx=ddx(x2+2)2.sin⁡x =sin⁡xddx(x2+2)2×(x2+2)2ddxsin⁡x  \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 (x2+2)2{\left( {{x^2} + 2} \right)^2} then, remove, power further, will some.
dydx=sin⁡x(x2+2)ddx(x2+2)+(x2+2)2cos⁡x dydx=sin⁡x(x2+2)[ddx(x2)+ddx(2)]+(x2+2)2cos⁡x dydx=sin⁡x(x2+2)[2x+0]+(x2+2)2cos⁡x  \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 \\\ ( ∵(2)\because (2) is constant we cannot do differentiation)
dydx=2(x2+2).(2x+0).sin⁡x+(x2+2)2.cos⁡x\dfrac{{dy}}{{dx}} = 2\left( {{x^2} + 2} \right).\left( {2x + 0} \right).\sin x + {\left( {{x^2} + 2} \right)^2}.\cos x
⇒ 4x(x2+2)sin⁡x+(x2+2)2cos⁡x\Rightarrow \,4x\left( {{x^2} + 2} \right)\sin x + {\left( {{x^2} + 2} \right)^2}\cos x
Taking common (x2+2)\left( {{x^2} + 2} \right)on both sides, we get
⇒(x2+2)  (4xsin⁡x+(x2+2)cos⁡x)\Rightarrow \left( {{x^2} + 2} \right)\,\,\left( {4x\sin x + \left( {{x^2} + 2} \right)\cos x} \right)
dydx=(x2+2)  (4xsin⁡x+(x2+2)cos⁡x)\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 [f(x)g(x)]\left[ {f\left( x \right)g\left( x \right)} \right]when they differentiate this value with respect to x{\text{x}} then
ddx[f(x)g(x)]=g(x)ddx[f(x)]+f(x)ddx[g(x)]\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]