Question

Find the second order derivatives of x^2 log x

Answer 1

Step-by-step explanation:

ans of given question is given

Answer 2

Answer:

The second order derivatives of $x^{2} logx$ is $3 + logx^{2}$.

Step-by-step explanation:

Differentiation formula of (u v)' = u' v + v' u

Derivative of (u v) is $\frac{d}{dx} (uv)= v \frac{d}{dx} (u) + u \frac{d}{dx}(v)$

Differentiation of $x$ is $\frac{dx}{dx}$ or (x)' = 1

Derivative of $logx$ is $\frac{1}{x}$

Derivative of $x^{2}$ is $2x$

Logarithm property $logx^{n} =nlogx$

Given $y = x^{2} logx$

Differentiate with respect to $x$

$\frac{dy}{dx} = \frac{d}{dx} (x^{2} logx)\\ \frac{dy}{dx} = x^{2} \frac{d}{dx} + logx\frac{d}{dx} (x^{2} )\\\frac{dy}{dx} = x^{2} \frac{1}{x} +logx (2x)$

Therefore,

$\frac{dy}{dx} = x+2xlogx$

Now again differentiate with respect to $x$ we get

$\frac{d^{2} y}{dx^{2} }= \frac{d}{dx}(x+2xlogx)\\ \frac{d^{2} y}{dx^{2} }=\frac{d}{dx} (x)+2\frac{d}{dx}(xlogx)\\ = 1+2[x.\frac{d}{dx}(logx)+logx.\frac{d}{dx}(x)\\ =1+2[x(\frac{1}{x}+logx.(1)]\\ =1+2(1+logx)$

Therefore, the second order derivatives of $x^{2} logx$ is

$\frac{d^{2} y}{dx^{2} } = 1+2+2logx\\\frac{d^{2}y }{dx^{2} } = 3+logx^{2}$