If x^m y^n=(x+y)^m+n, then dydx is:

Question

If $x^m\cdot y^n=(x+y)^{m+n}$, then $\dfrac{dy}{dx}$ is:

Accepted Answer

$${x^m}.{y^n} = {\left( {x + y} \right)^{m + n}}$$

$$ \Rightarrow m\ln x + n\ln y = \left( {m + n} \right)\ln \left( {x + y} \right)$$

Differentiating both sides.

$$\therefore$$ $${m \over x} + {n \over y}{{dy} \over {dx}} = {{m + n} \over {x + y}}\left( {1 + {{dy} \over {dx}}} \right)$$

$$ \Rightarrow \left( {{m \over x} - {{m + n} \over {x + y}}} \right) = \left( {{{m + n} \over {x + y}} - {n \over y}} \right){{dy} \over {dx}}$$

$$ \Rightarrow {{my - nx} \over {x\left( {x + y} \right)}} = \left( {{{my - nx} \over {y\left( {x + y} \right)}}} \right){{dy} \over {dx}}$$

$$ \Rightarrow {{dy} \over {dx}} = {y \over x}$$

Solutions · 1