Question

tell me the derivatieve of tan2x with the solution....

Answer 1

sec^2 2x . 2 =2/cos ^2x

Answer 2

Answer:

The answer is: 2sec22x

Explanation:

f(x+h)=tan2(x+h)
f(x)=tan2x
By first principles,
f'(x)=f(x+h)−f(x)h
Now, I will not put limh→0 on every line, you must remember to put it

=tan2(x+h)−tan2xh
=tan(2x+2h)−tan2xh
Expanding using tan(a+b)=tana+tanb1−tanatanb

= tan2x+tan2h1−tan2xtan2h−tan2xh
=tan2x+tan2h−tan2x+tan22xtan2hh(1−tan2xtan2h)
=tan2h(1+tan22x)h(1−tan2xtan2h)
Multiplying and dividing by two, and 1+tan22x=sec22x,
=tan2h⋅2sec22x2h(1−tan2xtan2h)
Now, let us analyse the tan(2h)2h part,
=sin(2h)2h⋅(1cos(2h))
As h→0,
sin(2h)2h=1
So,
tan(2h)2h=1cos(2h)
Putting this back in the equation,
=2sec22xcos2h(1−tan2xtan2h)
As, h→0, cos2h=1,tan2h=0
Putting the above eqn.,
=2sec22x(Don't put limh→0 in this line)