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JEE Main 2019
Differentiation
Differentiation
Medium

Question

If xm.yn=(x+y)m+n,{x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}, then dydx{{{dy} \over {dx}}} is

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Solution

xm.yn=(x+y)m+n{x^m}.{y^n} = {\left( {x + y} \right)^{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) 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) (mxm+nx+y)=(m+nx+yny)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}} mynxx(x+y)=(mynxy(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}} dydx=yx \Rightarrow {{dy} \over {dx}} = {y \over x}

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