Question
If $\text{xy}=4,$ prove that $\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
✓
Answer
We have, $\text{xy}=4$
$\Rightarrow\text{y}=\frac{4}{\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\frac{4}{\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4(-1\times\text{x}^{-1-1})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4}{\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4}{\big(\frac{4}{\text{y}}\big)^2}\ \Big[\because\text{x}=\frac{4}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4\text{y}^2}{16}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2}{4}$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}+4\text{y}^2=3\text{y}^2$
$\Rightarrow4\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}^2$
Dividing both side by x,
$\Rightarrow\frac{4}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
$\Rightarrow\text{y}=\frac{4}{\text{x}}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\frac{4}{\text{x}}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\frac{\text{d}}{\text{dx}}\big(\text{x}^{-1}\big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4(-1\times\text{x}^{-1-1})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\Big(-\frac{1}{\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4}{\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4}{\big(\frac{4}{\text{y}}\big)^2}\ \Big[\because\text{x}=\frac{4}{\text{y}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{4\text{y}^2}{16}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}^2}{4}$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}=-\text{y}^2$
$\Rightarrow4\frac{\text{dy}}{\text{dx}}+4\text{y}^2=3\text{y}^2$
$\Rightarrow4\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}^2$
Dividing both side by x,
$\Rightarrow\frac{4}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=\frac{3\text{y}^2}{\text{y}}$
$\Rightarrow\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big)=3\text{y}$
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