Find derivative of sin2x,cos2x and tan2x using first principle
Let us assume
f (x) = sin2x
f(x + h) = sin2(x + h)
Now as we know derivative is
d(f(x))/dx = f'(x) = lim h→0 [[sin2(x + h) - sin 2x] / [x + h - x]]
Now we know sin A - sin B =2 sin ((A-B)/2) * cos ((A+B)/2)
Thus , f'(x) = lim h→0 [[2sin((2(x + h) - 2x)/2) * cos((2(x+h) + 2x)/2)] / [x + h - x]]
= lim h→0 [[2sin(2h/2) * cos((4x+2h)/2)] / [x + h - x]]
= lim h→0 [[2sin(h) * cos(2x+h)] / h
=2 cos 2x lim h→0 [sin(h)] / h
= 2 cos 2x * 1 = 2 cos2x
Similarly do for others or if still you do not get your answer, you can ask the same in next thread.
f (x) = sin2x
f(x + h) = sin2(x + h)
Now as we know derivative is
d(f(x))/dx = f'(x) = lim h→0 [[sin2(x + h) - sin 2x] / [x + h - x]]
Now we know sin A - sin B =2 sin ((A-B)/2) * cos ((A+B)/2)
Thus , f'(x) = lim h→0 [[2sin((2(x + h) - 2x)/2) * cos((2(x+h) + 2x)/2)] / [x + h - x]]
= lim h→0 [[2sin(2h/2) * cos((4x+2h)/2)] / [x + h - x]]
= lim h→0 [[2sin(h) * cos(2x+h)] / h
=2 cos 2x lim h→0 [sin(h)] / h
= 2 cos 2x * 1 = 2 cos2x
Similarly do for others or if still you do not get your answer, you can ask the same in next thread.