Answer:
[tex]\qquad\qquad\boxed{ \sf{ \: \: \bf \:\dfrac{ {d}^{2}}{d {x}^{2} } ( {x}^{2}logx ) = 3 + 2logx \: }} \\ \\ [/tex]
Step-by-step explanation:
Let assume that
[tex]\sf \:y = {x}^{2}logx \\ \\ [/tex]
On differentiating both sides w. r. t. x, we get
[tex]\sf \:\dfrac{d}{dx}y =\dfrac{d}{dx}({x}^{2}logx)\\ \\ [/tex]
[tex]\sf \:\dfrac{dy}{dx} = {x}^{2}\dfrac{d}{dx}logx + logx\dfrac{d}{dx} {x}^{2} \\ \\ [/tex]
[tex]\sf \:\dfrac{dy}{dx} = {x}^{2} \times \dfrac{1}{x} + logx(2x) \\ \\ [/tex]
[tex]\sf \:\dfrac{dy}{dx} = x + 2xlogx \\ \\ [/tex]
[tex]\sf \:\dfrac{dy}{dx} = x(1 + 2logx) \\ \\ [/tex]
On differentiating both sides w. r. t. x, we get
[tex]\sf \:\dfrac{ {d}^{2} y}{d {x}^{2} } = \dfrac{d}{dx}[ x(1 + 2logx)] \\ \\ [/tex]
[tex]\sf \:\dfrac{ {d}^{2} y}{d {x}^{2} } = x\dfrac{d}{dx}(1 + 2logx) + (1 + 2logx)\dfrac{d}{dx}x \\ \\ [/tex]
[tex]\sf \:\dfrac{ {d}^{2} y}{d {x}^{2} } = x \times \left(0 + 2 \times \dfrac{1}{x} \right) + 1 + 2logx \\ \\ [/tex]
[tex]\sf \:\dfrac{ {d}^{2} y}{d {x}^{2} } = x \times \left(\dfrac{2}{x} \right) + 1 + 2logx \\ \\ [/tex]
[tex]\sf \:\dfrac{ {d}^{2} y}{d {x}^{2} } = 2 + 1 + 2logx \\ \\ [/tex]
[tex]\bf\implies \: \bf \:\dfrac{ {d}^{2} y}{d {x}^{2} } = 3 + 2logx \\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
[tex]\sf \:\dfrac{d}{dx}(uv) = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \\ \\ [/tex]
[tex]\sf \:\dfrac{d}{dx}logx = \dfrac{1}{x} \\ \\ [/tex]
[tex]\sf \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \\ \\ [/tex]
[tex]\sf \:\dfrac{d}{dx} k = 0 \\ \\ [/tex]
[tex]\sf \:\dfrac{d}{dx} k \: f(x) = k \: \dfrac{d}{dx}f(x) \\ \\ [/tex]
