CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\},$ find $\frac{\text{dy}}{\text{dx}}.$
✓
Answer
Here, $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\}$
Let $\text{x}=\cos\theta, \text{So},$
$\text{y}=\cos^{-1}\Big\{\frac{2\cos\theta-3\sqrt{1-\cos^2\theta}}{\sqrt{13}}\Big\}$
$=\cos^{-1}\Big\{\frac{2}{\sqrt{13}}\cos\theta-\frac{3}{13}\sin\theta\Big\}$
Let $\cos\phi=\frac{2}{\sqrt{13}}$
$\Rightarrow\sin\phi=\sqrt{1-\cos^2\phi}$
$=\sqrt{1-\Big(\frac{2}{\sqrt{13}}\Big)^2}$
$=\sqrt{\frac{13-4}{13}}$
$=\sqrt{\frac{9}{13}}$
$\sin\phi=\frac{3}{\sqrt{13}}$
So,
$\text{y}=\cos^{-1}\big\{\cos\phi\cos\theta-\sin\phi\sin\theta\big\}$
$=\cos^{-1}\big[\cos(\theta+\phi)\big]$
$\text{y}=\phi+\theta$
$\text{y}=\cos^{-1}\Big(\frac{2}{\sqrt{13}}\Big)+\cos^{-1}\text{x}$
$\Big[\text{Since, x}=\cos\theta,\cos\phi=\frac{2}{\sqrt{13}}\Big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0+\Big(-\frac{1}{\sqrt{1-\text{x}^2}}\Big)$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
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