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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
Let $\sin\text{y}=\text{x}\sin(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is :
  • A
    $\frac{\sin\text{a}}{\sin\text{a}\sin^2(\text{a}+\text{y})}$
  • ✓
    $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
  • C
    $\sin\text{a}\sin^2(\text{a}+\text{y})$
  • D
    $\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
✓

Answer

Correct option: B.
$\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
We have, $\sin\text{y}=\text{x}\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})\big]$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\big\{\sin(\text{a}+\text{y})\big\}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})\times1+\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})+\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}-\text{x}\cos(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\big\{\cos\text{y}-\text{x}\cos(\text{a}+\text{y})\big\}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\Big\{\cos\text{y}-\frac{\sin\text{y}}{\sin(\text{a}+\text{y})}\times\cos(\text{a}+\text{y})\Big\}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\begin{bmatrix} \because \sin\text{y}=2\sin\text{c}\cos\text{x} \\ \therefore\text{x}=\frac{\sin\text{y}}{\sin(\text{a}+\text{y})} \end{bmatrix}$
$\Rightarrow\Big\{\frac{\sin(\text{a}+\text{y})\cos\text{y}-\sin\text{y}\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\sin(\text{a}+\text{y}-\text{y})}{\sin(\text{a}+\text{y})}\times\frac{\text{dy}}{\text{dx}}=\sin(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$

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