CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $x^{13} y^7 = (x + y)^{20}$, prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
✓
Answer
Here,
$x^{13}y^7 = (x + y)^{20}$
Taking log on both the sides,
$\log(\text{x}^{13}\text{y}^7)=\log(\text{x}+\text{y})^{20}$
$13\log\text{x}+7\log\text{y}=20\log(\text{x}+\text{y})$
$\big[\text{Since},\log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule,
$13\frac{\text{d}}{\text{dx}}(\log\text{x})+7\frac{\text{d}}{\text{dx}}(\log\text{y})=20\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{\text{x}+\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})$
$\frac{13}{\text{x}}+\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{20}{(\text{x}+\text{y})}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\frac{7}{\text{y}}\frac{\text{dy}}{\text{dx}}-\frac{20}{(\text{x}+\text{y})}=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{7}}{\text{y}}-\frac{20}{(\text{x}+\text{y})}\Big]=\frac{20}{(\text{x}+\text{y})}-\frac{13}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{2(\text{x}+\text{y})-20\text{y}}{\text{y}(\text{x}+\text{y})}\Big]=\Big[\frac{20\text{x}-13(\text{x}+\text{y})}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=\Big[\frac{20\text{x}-13\text{x}-13\text{y}}{\text{x}(\text{x}+\text{y})}\Big]\Big(\frac{\text{y}(\text{x}+\text{y})}{7\text{x}+7\text{y}-20\text{y}}\Big)$
$=\frac{\text{y}}{\text{x}}\Big(\frac{7\text{x}-13\text{y}}{7\text{x}-13\text{y}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
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