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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
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Answer

We have, $\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiate with respect to y,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\{\sin(\text{a}+\text{y})\}+\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule and chain rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}+\sin(\text{a}+\text{y})(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\{1-\text{x}\cos(\text{a}+\text{y})\}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\text{x}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y}){-\text{y}\cos(\text{a}+\text{y})}}$
Hence, proved.

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