Question

Find the second order derivatives of the function.x^3.logx

Answer 1

Let $\bf{y=x^3.logx}$
differentiate y with respect to x
$\bf{\frac{dy}{dx}=x^3.\frac{d(logx)}{dx}+logx.\frac{d(x^3)}{dx}}\\\\=\bf{x^3.\frac{1}{x}+logx.(3x^{3-1})}\\\\=\bf{x^2+3x^2.logx}$

so,$\bf{\frac{dy}{dx}=x^3+3x^2logx}$
now differentiate $\frac{dy}{dx}$ once again,
$\bf{\frac{d^2y}{dx^2}=\frac{d(x^2)}{dx}+\frac{d(3x^2logx)}{dx}}\\\\=\bf{2x+3x^2\frac{d(logx)}{dx}+logx.\frac{d(3x^2)}{dx}}\\\\=\bf{2x+3x^2.\frac{1}{x}+logx.6x^{2-1}}\\\\=\bf{2x+3x+6x.logx}$

hence, d²y/dx² = 5x + 6xlogx

Answer 2

$y = {x}^{3} log(x) \\ \frac{dy}{dx} = 3 {x}^{2} log(x) + {x}^{3} {(\frac{1}{x} )} \\ \frac{dy}{dx} = 3 {x}^{2} log(x) + {x}^{2} \\ now \: again \: diffrentiating \: wrt \: x \\ \frac{ {d}^{2} y}{d {x}^{2} } = 6x log(x) + 3 {x}^{2} ( \frac{1}{x} ) + 2x \\ \frac{ {d}^{2}x }{d {y}^{2} } = 6x log(x) + 3x + 2x \\ \frac{ {d}^{2}x }{d {y}^{2} } = 6 xlog(x) + 5x$