A) \[\frac{1}{2}{{\sec }^{2}}x\]
B) \[{{\sec }^{2}}x\]
C) \[\sec x\tan x\]
D) \[{{e}^{1/2\log (1+{{\tan }^{2}}x)}}\]
E) \[{{e}^{1/2\log (1+{{\tan }^{2}}x)}}.\frac{1}{2}\frac{1}{(1+{{\tan }^{2}}x)}\]
Correct Answer: C
Solution :
\[y={{e}^{(1/2)\log (1+{{\tan }^{2}}x)}}\] \[y={{({{\sec }^{2}}x)}^{1/2}}=\sec x\] On differentiating, we get \[\frac{dy}{dx}=\sec x\tan x\]You need to login to perform this action.
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