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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\sqrt{\sin\text{x}+\text{y}},$ then $\frac{\text{dy}}{\text{dx}}$ equals :
  • A
    $\frac{\cos\text{x}}{2\text{y}-1}$
  • ✓
    $\frac{\cos\text{x}}{1-2\text{y}}$
  • C
    $\frac{\sin\text{x}}{1-2\text{y}}$
  • D
    $\frac{\sin\text{x}}{2\text{y}-1}$
✓

Answer

Correct option: B.
$\frac{\cos\text{x}}{1-2\text{y}}$
$\text{y}=\sqrt{\sin\text{x}+\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\times\Big(\cos\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos}{2\sqrt{\sin\text{x}+\text{y}}}+\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\frac{\text{dy}}{\text{dx}}=\frac{\cos}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\sqrt{\sin\text{x}+\text{y}}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\sqrt{\sin\text{x}+\text{y}}}$
$\Big(1-\frac{1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\Big(\frac{2\text{y}-1}{2\text{y}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{2\text{y}-1}$

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