If x^m.y^n=(x+y)^ m+n , prove that: dy dx = y x

Question

If $\text{x}^\text{m}.\text{y}^\text{n}=(\text{x}+\text{y})^{\text{m}+\text{n}},$ prove that:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$

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Answer

We have,
$\text{x}^\text{m}.\text{y}^\text{n}=(\text{x}+\text{y})^{\text{m}+\text{n}}\ \ \dots(\text{i})$
Differentiating Eq. (i) w.r.t. x, we get
$\frac{\text{d}}{\text{dx}}(\text{x}^\text{m}.\text{y}^\text{n})=\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})^{\text{m}+\text{n}}$
$\Rightarrow\ \text{x}^\text{m}.\frac{\text{d}}{\text{dy}}\text{y}^\text{n}.\frac{\text{dy}}{\text{dx}}+\text{y}^\text{n}.\frac{\text{d}}{\text{dx}}\text{x}^\text{m}$ $=(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\Rightarrow\ \text{x}^\text{m}.\text{ny}^{\text{n}-1}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{n}.\text{mx}^{\text{m}-1}$ $=(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[\text{x}^\text{m}.\text{ny}^{\text{n}-1}-(\text{m}+\text{n}).(\text{x}+\text{y})^{\text{m}+\text{n}-1}\big]$ $=(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}-\text{y}^\text{n}\text{mx}^{\text{m}-1}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[\text{nx}^\text{m}\text{y}^{\text{n}-1}-(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}-1}\big]$ $=(\text{m}+\text{n}(\text{x}+\text{y})^{\text{m}+\text{n}-1}-\frac{\text{y}^{\text{n}-1}.\text{y}.\text{mx}^{\text{m}-1}}{\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}}{(\text{x}-\text{y})}-\frac{\text{y}^{\text{n}-1}.\text{y}.\text{mx}^\text{m}}{\text{x}}}{\frac{\text{nx}^\text{m}\text{y}^\text{n}}{\text{y}}-(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}\frac{1}{(\text{x}+\text{y})}}$
$=\frac{\frac{\text{x}(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}-(\text{x}+\text{y}).\text{y}.^{\text{n}-1}\text{y}.\text{mx}^\text{m}}{(\text{x}+\text{y}).\text{x}}}{\frac{(\text{x}+\text{y})\text{nx}^\text{m}\text{y}^\text{n}-\text{y}(\text{m}+\text{n})(\text{x}+\text{y})^{\text{m}+\text{n}}}{(\text{x}+\text{y}).\text{y}}}$
$=\frac{\frac{\text{x}(\text{m}+\text{n})\text{x}^\text{m}.\text{y}^\text{n}-\text{m}(\text{x}+\text{y})\text{y}^\text{n}\text{x}^\text{m}}{(\text{x}+\text{y}).\text{x}}}{\frac{(\text{x}+\text{y})\text{nx}^\text{m}.\text{y}^\text{n}-\text{y}(\text{m}+\text{n}).\text{x}^\text{m}.\text{y}^\text{n}}{(\text{x}+\text{y}).\text{y}}}$ $\big[\because(\text{x}+\text{y})^{\text{m}+\text{n}}=\text{x}^\text{m}.\text{y}^\text{n}\big]$
$=\frac{\text{x}^\text{m}\text{y}^\text{n}[\text{mx}+\text{nx}-\text{mx}-\text{my}].(\text{x}+\text{y})\text{y}}{\text{x}^\text{m}\text{y}^\text{n}[\text{nx}+\text{ny}-\text{my}-\text{ny}].(\text{x}+\text{y})\text{x}}$
$=\frac{\text{y}}{\text{x}}\ \ \dots(\text{i})$
Hence proved.

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