Question #47074

If y=tan2x

2sec^22x
2sec2x
sec^2
sec2x

Expert's answer

47074, Math, Differential Calculus

If y=tan2xy = \tan 2x

Solution:


y=(tan2x)=1cos22x(2x)=2sec2xy' = (\tan 2x)' = \frac{1}{\cos^2 2x} \cdot (2x)' = 2 \sec 2x


Answer: 2 sec 2x

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