Question
If $\text{y}=\text{x}\sin(\text{a}+\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
✓
Answer
Here,
$\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiating with respect to x using the chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{dx}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})$
$(1-\text{x}\cos(\text{a}+\text{y}))\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{(1-\text{x}\cos(\text{a}+\text{y}))}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{\Big(1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})\Big)}\ \Big[\text{Since}\frac{\text{y}}{\sin(\text{a}+\text{y})}=\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
$\text{y}=\text{x}\sin(\text{a}+\text{y})$
Differentiating with respect to x using the chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\sin(\text{a}+\text{y})+\sin(\text{a}+\text{y})\frac{\text{dx}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}+\sin(\text{a}+\text{y})$
$(1-\text{x}\cos(\text{a}+\text{y}))\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{(1-\text{x}\cos(\text{a}+\text{y}))}$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin(\text{a}+\text{y})}{\Big(1-\frac{\text{y}}{\sin(\text{a}+\text{y})}\cos(\text{a}+\text{y})\Big)}\ \Big[\text{Since}\frac{\text{y}}{\sin(\text{a}+\text{y})}=\text{x}\Big]$
$\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin(\text{a}+\text{y})-\text{y}\cos(\text{a}+\text{y})}$
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