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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}=$
  • A
    2
  • B
    -2
  • C
    1
  • ✓
    -1
✓

Answer

Correct option: D.
-1
We have, $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)=\frac{1}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$

$\Rightarrow\cos(\text{x}+\text{y})+\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$

$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}-\cos(\text{x}+\text{y})$

$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$

$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$

$\Rightarrow\Big\{\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$

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