CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{y}=\cos^{-1}\Big\{\frac{2\text{x}-3\sqrt{1-\text{x}^2}}{\sqrt{13}}\Big\},$ find $\frac{\text{dy}}{\text{dx}}.$
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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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