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# what is the derivative of tan2x ? watch

1. what is the derivative of tan2x ?
2. 2sec^2(2x)
3. let u=2x
4. express y = tan2x as y = sin2x / cos2x and use quotient rule.

You can let u = 2x and work without 2x if it will make the life easier for you. But don't forget it in the final answer.
5. Re: what is the derivative of tan2x ?
express y = tan2x as y = sin2x / cos2x and use quotient rule.

You can let u = 2x and work without 2x if it will make the life easier for you. But don't forget it in the final answer.
I used that method and I got (2(cos2x)^2-2(sin2x)^2)/(cos2x)^2, The actual question in the book is to find dy/dt of the parametric equations t=tanx and y=tan2x when x= 30 degrees, my final gradient was -3 but the book showed it to be 6 i thought my error was in the differentiation of tan2x
6. (Original post by liam251)
I used that method and I got (2(cos2x)^2-2(sin2x)^2)/(cos2x)^2, The actual question in the book is to find dy/dt of the parametric equations t=tanx and y=tan2x when x= 30 degrees, my final gradient was -3 but the book showed it to be 6 i thought my error was in the differentiation of tan2x
The - in the numerator should be a +.
7. The - in the numerator should be a +.
oooooooooohhhhhhhhhhhhh thanks I knew there was a silly error in that differentiation of tan2x
8. y=tan f(x) implies that =

So

You can also prove this result by setting tan2x equal to sin2x/cos2x.

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