Question

what is double derivative of x^m*y^n=x+y^m+n​

Answer 1

Solution:-

$\rm \to \: {x}^{m} y^{n} = (x + y)^{m + n}$

=> Taking Both side log

$\rm \to \: log( {x}^{m} {y}^{n} ) = log(x + y)^{m + n}$

=> Using Log properties

$\rm \to \: log x ^{y} = y logx$

=> We can write

$\rm \to \: log( {x}^{m} {y}^{n} ) = (m + n) log(x + y)$

$\rm \to \: log(x ^{m} ) + log( {y}^{n} ) = (m + n) log(x + y)$

$\rm \to \: m log(x) + n log(y) = (m + n) log(x + y)$

=> Now differentiate w.r.t x

$\rm \to \: m \dfrac{1}{x} + n \dfrac{1}{y} \dfrac{dy}{dx} = (m + n) \dfrac{1}{x + y} \bigg(1 + \dfrac{dy}{dx} \bigg)$

$\rm \to \: \dfrac{m}{x} + \dfrac{n}{y} \dfrac{dy}{dx} = \dfrac{m + n}{x + y} \bigg(1 + \dfrac{dy}{dx} \bigg)$

$\rm \to \: \dfrac{m}{x} + \dfrac{n}{y} \dfrac{dy}{dx} = \dfrac{m + n}{x + y} + \bigg( \dfrac{m + n}{x + y} \bigg) \dfrac{dy}{dx}$

=> Now take dy/dx on one side

$\rm \to \: \dfrac{n}{y} \dfrac{dy}{dx} - \dfrac{m + n}{x + y} \dfrac{dy}{dx} = \dfrac{m + n}{x + y} - \dfrac{m}{x}$

$\rm \to \: \bigg( \dfrac{n}{y} - \dfrac{m + n}{x + y} \bigg) \dfrac{dy}{dx} = \dfrac{(m + n)x - m(x + y)}{x(x + y)}$

$\rm \to \: \bigg(\dfrac{(x + y)n - (m + n)y}{y(x + y)} \bigg) \dfrac{dy}{dx} = \dfrac{mx + nx - mx - my}{x(x + y)}$

$\rm \to \: \bigg(\dfrac{(x + y)n - (m + n)y}{y \cancel{(x + y)} } \bigg) \dfrac{dy}{dx} = \dfrac{ \cancel{mx }+ nx - \cancel{mx} - my}{x \cancel{(x + y)}}$

$\rm \to \: \bigg( \dfrac{nx + ny - my - ny}{y} \bigg) \dfrac{dy}{dx} = \dfrac{nx - my}{x}$

$\rm \to \: \bigg( \dfrac{nx + \cancel{ny} - my - \cancel{ ny}}{y} \bigg) \dfrac{dy}{dx} = \dfrac{nx - my}{x}$

$\rm \to \bigg( \dfrac{nx - my}{y} \bigg) \dfrac{dy}{dx} = \dfrac{nx - my}{x}$

$\rm \to \bigg( \dfrac{ \cancel{nx - my}}{y} \bigg) \dfrac{dy}{dx} = \dfrac{ \cancel{nx - my}}{x}$

$\rm \to \bigg( \dfrac{1}{y} \bigg) \dfrac{dy}{dx} = \dfrac{1}{x}$

$\rm \to \: \dfrac{dy}{dx} = \dfrac{y}{x}$

=> again differentiate w.r.t x

$\rm \to \: \dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{ \dfrac{dy}{dx}x - y }{ {x}^{2} }$

=>Put the value of dy/dx

$\rm \to \: \dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{ \dfrac{y}{x} \times x - y }{ {x}^{2} }$

$\rm \to \: \dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{y - y}{ {x}^{2} } = \dfrac{0}{ {x}^{2} } = 0$

Answer

$\rm \to \: \dfrac{ {d}^{2}y }{d {x}^{2} } = 0$