Question

If (x + y) ^(m + n) = x^m . y^n then prove that . dy/dx = y/x.

Answer 1

Question :-

$\red{ \star \: if \: \: \: {x}^{m} {y}^{n} = {(x + y)}^{(m + n)} } \\ \green{ then \: show \: that \: \boxed{\frac{dy}{dx} = \frac{y}{x} }}$

Step - by - step explanation:-

Used properties :-

$\implies \: \boxed{ log( {m}^{n} ) = n log(m) } \\ \\ \implies \: \boxed{ log(mn) = log(m) + log(n) }$

Solution :-

According to the question,

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

Taking logarithm of both sides ,

$log( {x}^{m} {y}^{n} ) = log \big( {(x + y)}^{(m + n)} \big)$

Using the given properties of logarithm,

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

Now differentiate with respect to" x " of both sides ,

$\because \: \red{ \frac{d( log(x)) }{dx} = \frac{1}{x} } \\ \\ \therefore \: \frac{m}{x} + \frac{n}{y} \: \frac{dy}{dx} = \frac{(m + n)}{(x + y)} \times (1 + \frac{dy}{dx} ) \\ \\ \frac{(m + n)}{(x + y)} \: \frac{dy}{dx} - \frac{n}{y} \: \frac{dy}{dx} \: = \frac{m}{x} - \frac{(m + n)}{(x + y)} \\ \\ \frac{dy}{dx} \bigg( \frac{(m + n)y - n(x + y)}{(x + y)y} \bigg) = \frac{mx + my - mx - nx}{x(x + y)} \\ \\ \frac{dy}{dx} \bigg( \frac{my - nx}{(x + y)y} \bigg) = \frac{my - nx}{x(x + y)} \\ \\ \therefore \: \\ \\ \implies \: \green{\boxed{ \red{\boxed{ \frac{dy}{dx} = \frac{y}{x} }}}}$

Hence proved.