CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0,$ prove that $(1+\text{x})^2\frac{\text{dx}}{\text{dx}}+1=0$
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Answer
We have $\text{x}\sqrt{1+\text{y}}+\text{y}\sqrt{1+\text{x}}=0$ $\Rightarrow\text{x}\sqrt{1+\text{y}}=-\text{y}\sqrt{1+\text{x}}$ Squaring both sides, we get, $\Rightarrow\big(\text{x}\sqrt{1+\text{y}}\big)^2=(-\text{y}\sqrt{1+\text{x}}\big)^2$ $\Rightarrow\text{x}^2\big(1+\text{y}\big)=\text{y}^2(1+\text{x}\big)$ $\Rightarrow\text{x}^2+\text{x}^2\text{y}=\text{y}^2+\text{y}^2\text{x}$ $\Rightarrow\text{x}^2-\text{y}^2=\text{y}^2\text{x}-\text{x}^2\text{y}$ $\Rightarrow(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}\text{y}(\text{y}-\text{x})$ $\Rightarrow(\text{x}+\text{y})=-\text{x}\text{y}$ $\Rightarrow\text{y}+\text{x}\text{y}=-\text{x}$ $\Rightarrow\text{y}(1+\text{x})=-\text{x}$ $\Rightarrow\text{y}=\frac{-\text{x}}{(1+\text{x})}$ Differentiating with respect to x, we get $\Rightarrow\frac{\text{dy}}{\text{dx}}=\bigg[\frac{-(1+\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-(-\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+1)}{(1+\text{x})^2}\bigg]$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-(1+\text{x})(1)+\text{x}(1)}{(1+\text{x})^2}\Big]$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{-1-\text{x}+\text{x}}{(1+\text{x})^2}\Big]$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-1}{(1+\text{x})^2}$ $\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}=-1$ $\Rightarrow(1+\text{x})^2\frac{\text{dy}}{\text{dx}}+1=0$
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