Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
If xy2 = 1, prove that $2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
✓
Answer
We have xy2 = 1 .....(i) Differentiating with respect to x, we get, $\frac{\text{d}}{\text{dx}}(\text{xy}^2)=\frac{\text{d}}{\text{dx}}(1)$ $\Rightarrow\text{x}\frac{\text{d}}{\text{dx}}(\text{y}^2)+\text{y}^2\frac{\text{d}}{\text{dx}}(\text{x})=0$ $\Rightarrow\text{x}(2\text{y})\frac{\text{d}}{\text{dx}}+\text{y}^2(1)=0$ $\Rightarrow2\text{xy}\frac{\text{dy}}{\text{dx}}=-\text{y}^2$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{2\text{xy}}$ $\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\text{x}}$ Put $\text{x}=\frac{1}{\text{y}^2}$ from equation (i) $ \Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-\text{y}}{2\Big(\frac{1}{\text{y}^2}\Big)}$ $\Rightarrow2\frac{\text{dy}}{\text{dx}}=-\text{y}^3$ $\Rightarrow2\frac{\text{dy}}{\text{dx}}+\text{y}^3=0$
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