If $y=x^{x}$, prove that $\dfrac{d^{2}y}{dx^{2}}-\dfrac{1}{y}\!\left(\dfrac{dy}{dx}\right)^{2}-\dfrac{y}{x}=0$.

CBSE Class 12 · 2016 · DifferentiationLong Answer

y = x^{x}

, prove that

\frac{d ^{2} y}{d x ^{2}} - \frac{1}{y} (\frac{d y}{d x})^{2} - \frac{y}{x} = 0

Answer: Proved

Take logarithm:

lo g y = x lo g x

. Differentiate:

\frac{1}{y} \frac{d y}{d x} = lo g x + 1 \Rightarrow \frac{d y}{d x} = y (1 + lo g x)

Differentiate again (product rule):

\frac{d ^{2} y}{d x ^{2}} = \frac{d y}{d x} (1 + lo g x) + y \cdot \frac{1}{x}

Replace

(1 + lo g x) = \frac{1}{y} \frac{d y}{d x}

in the first term:

\frac{d ^{2} y}{d x ^{2}} = \frac{1}{y} (\frac{d y}{d x})^{2} + \frac{y}{x}

\frac{d ^{2} y}{d x ^{2}} - \frac{1}{y} (\frac{d y}{d x})^{2} - \frac{y}{x} = 0

CBSE Class 12 Board · 2016 · Subjective