Verified Solution • 2025–26 Session moderate QJEEM06Ma422c: JEE Main 2006 Mathematics - Class 12 JEEM06Ma422c Solution: If then is If xm.yn=(x+y)m+n,{x^m}.{y^n} = {\left( {x + y} \right)^{m + n}},xm.yn=(x+y)m+n, then dydx{{{dy} \over {dx}}}dxdy is Multiple Choice Options A yx{y \over x}xy B x+yxy{{x + y} \over {xy}}xyx+y C xyxyxy D xy{x \over y}yx Step-by-Step Verified Solution xm.yn=(x+y)m+n{x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}xm.yn=(x+y)m+n ⇒mlnx+nlny=(m+n)ln(x+y)\Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)⇒mlnx+nlny=(m+n)ln(x+y) Differentiating both sides. ∴\therefore∴ mx+nydydx=m+nx+y(1+dydx){m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)xm+yndxdy=x+ym+n(1+dxdy) ⇒(mx−m+nx+y)=(m+nx+y−ny)dydx\Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}⇒(xm−x+ym+n)=(x+ym+n−yn)dxdy ⇒my−nxx(x+y)=(my−nxy(x+y))dydx\Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}⇒x(x+y)my−nx=(y(x+y)my−nx)dxdy ⇒dydx=yx\Rightarrow {{dy} \over {dx}} = {y \over x}⇒dxdy=xy Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring