A) \[\frac{\sqrt{3-{{\sin }^{2}}x}}{\cos y}\]
B) \[-\frac{\sqrt{3-{{\sin }^{2}}x}}{\cos y}\]
C) \[\cos y\sqrt{3-{{\sin }^{2}}x}\]
D) \[-\cos y\sqrt{3-{{\sin }^{2}}x}\]
Correct Answer: B
Solution :
| Differentiating the given equation with respect to x, we get \[\frac{d}{dx}\left[ \int\limits_{\frac{\pi }{3}}^{x}{\sqrt{3-{{\sin }^{2}}t}dt} \right]+\frac{d}{dx}\left[ \int\limits_{0}^{y}{\cos \,t\,\,dt} \right]=0\] |
| \[\Rightarrow \,\,\sqrt{3-{{\sin }^{2}}x}.\frac{d}{dx}x-\sqrt{3-{{\sin }^{2}}\frac{\pi }{3}}\,\,\frac{d}{dx}\left( \frac{\pi }{3} \right)\] \[+\cos y\frac{dy}{dx}-\cos (0)\frac{d(0)}{dx}=0\] |
| \[\Rightarrow \,\,\,\sqrt{3-{{\sin }^{2}}x}-0+\cos y\frac{dy}{dx}-0=0\] |
| \[\Rightarrow \,\,\,\frac{dy}{dx}=-\frac{\sqrt{3-{{\sin }^{2}}x}}{\cos y}\] |
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