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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{x}^{3}+\text{x}^{2}\text{y} +\text{x} \text{y}^{2}+\text{y}^{3} = 81$
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Answer

The given relationship is $\text{x}^{3}+\text{x}^{2}\text{y} +\text{x} \text{y}^{2}+\text{y}^{3} = 81$ differenting this relationship with respect to x, we obtain $\frac{\text{d}}{\text{dx}}(\text{x}^{3}+\text{x}^{2}\text{y} +\text{x} \text{y}^{2}+\text{y}^{3}) =\frac{\text{d}}{\text{dx}}(81)$ $\Rightarrow\frac{\text{d}}{\text{dx}}(\text{x}^{3})+\frac{\text{d}}{\text{dx}}(\text{x}^{2}\text{y})+\frac{\text{d}}{\text{dx}}(\text{x}\text{y}^{2})+\frac{\text{d}}{\text{dx}}\text{(y}^{3}) =0$ $\Rightarrow 3\text{x}^{2}+\Big[\text{y}\frac{\text{d}}{\text{dx}}(\text{x}^{2})+\text{x}^{2}\frac{\text{dy}}{\text{dx}}\Big] +\Big[\text{y}^{2}\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{dy}}{\text{dx}}(\text{y}^{2})\Big]+3\text{y}^{2}\frac{\text{dy}}{\text{dx}}=0$$\Rightarrow3\text{x}^{2}+\Big[\text{y}.2\text{x}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{y}^{2}.1+\text{x}.2\text{y}.\frac{\text{dy}}{\text{dx}}\Big]+3\text{y}^{2}\frac{\text{dy}}{\text{dx}}= 0$
$\Rightarrow(\text{x}^{2} + 2\text{xy}+3\text{y}^{2})\frac{\text{dy}}{\text{dx}}+(3\text{x}^{2}+2\text{xy}+\text{y}^{2})= 0$$\therefore\frac{\text{dy}}{\text{dx}}= \frac{-(3\text{x}^{2}+2\text{xy}+\text{y}^{2})}{(\text{x}^{2}+2\text{xy}+3\text{y}^{2})}$

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