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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY2 Marks
Question
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{x}^{2}+\text{xy} + \text{y}^{2} =100$
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Answer

The given relationship is $\text{x}^{2}+\text{xy} + \text{y}^{2} =100$ differenting this relationship with respect to x, we obtain $\frac{\text{d}}{\text{dx}}(\text{x}^{2}+\text{xy} + \text{y}^{2}) = \frac{\text{d}}{\text{dx}}(100)$ $\frac{\text{d}}{\text{dx}}(\text{x}^{2})+\frac{\text{d}}{\text{dx}}(\text{xy})+\frac{\text{d}}{\text{dx}}\text{(y}^{2}) =0\ [\text{Derivative of constant function is 0}]$ $\Rightarrow 2\text{x}+ \Big[\text{y}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{dy}}{\text{dx}}\Big] +2\text{y} \frac{\text{dy}}{\text{dx}}=0\ \ [\text{Using product rule and chain rule]}$$\Rightarrow\ 2\text{x}+\text{y}.1+\text{x}.\frac{\text{dy}}{\text{dx}}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\ 2\text{x}+\text{y}+(\text{x}+2\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\therefore\frac{\text{dy}}{\text{dx}}= -\frac{2\text{x}+\text{y}}{\text{x} + 2\text{y}}$

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