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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\sin\text{y}=\text{x}\cos(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to :
  • ✓
    $\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
  • B
    $\frac{\cos\text{a}}{\cos^2(\text{a}+\text{y})}$
  • C
    $\frac{\sin^2\text{y}}{\cos\text{a}}$
  • D
    None of these.
✓

Answer

Correct option: A.
$\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
We have, $\sin\text{y}=\text{x}\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}\big[\text{x}\cos(\text{a}+\text{y})\big]$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=1\times\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{d}}{\text{dx}}(\text{a}+\text{y})$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})-\text{x}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\big[\cos\text{y}+\text{x}\sin(\text{a}+\text{y})\big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\Big[\cos\text{y}+\frac{\sin\text{y}}{\cos(\text{a}+\text{y})}\times\sin(\text{a}+\text{y})\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\begin{bmatrix}\because\sin\text{y}=\text{x}\cos(\text{a}+\text{y}) \\ \because\text{x}=\frac{\sin\text{y}}{\cos(\text{a}+\text{y})} \end{bmatrix}$
$\Rightarrow\Big[\frac{\cos(\text{a}+\text{y})\cos\text{y}+\sin\text{y}\sin(\text{a}+\text{y})}{\cos(\text{a}+\text{y})}\Big]\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\cos(\text{a}+\text{y}-\text{y})}{\cos(\text{a}+\text{y})}\times\frac{\text{dy}}{\text{dx}}=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$

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