Question
If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dx}}{\text{dx}}=\frac{\sin^2\text{y}}{(1-\text{x}\cos\text{y})}$
✓
Answer
Here,
$\text{y}=\text{x}\sin\text{y}$
Differentiate with respect to x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{1-\text{x}\cos\text{y}}$
$\text{y}=\text{x}\sin\text{y}$
Differentiate with respect to x,
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
[Using product rule]
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{1-\text{x}\cos\text{y}}$
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