If \(\rm x^m y^n =xya^{m+n}\), then what is \(\rm \frac{dy}{dx}\) equal to?

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)
• log ab = b log a 

Calculation:-  

\(\rm x^m y^n =xya^{m+n}\) , where a is constant 

Taking log both sides, we get

⇒ \(\rm \log (x^m y^n) =\log (xya^{m+n})\)

⇒ log x^m + log yⁿ = log x + log y + log a^m+n

⇒ m log x + n log y = log x + log y + (m + n) log a 

Differentiate both sides w.r.t  x, we get

⇒ \(\rm \frac{m}{x}+\frac{n}{y}\frac{\mathrm{d} y}{\mathrm{d} x}= \frac 1 x + \frac 1 y \frac {dy}{dx} + 0\)   [∵  \(\rm \frac{d \;\text{constant}}{dx} = 0\) ]  

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}\left [ \frac{n}{y}-\frac{1}{y} \right ]= \left [ \frac{1}{x}- \frac{m}{x} \right ]\) 

⇒  \(\rm \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-(m-1)y}{(n-1)x}\) . 

The correct option is 3.