Question

derivatives sin^2x from first principal​

Answer 1

Answer:

$2cos2x$

Step-by-step explanation:

The derivative of sin 2x has to be determined from first principles.

For a function f(x) the derivative from first principles is lim_(h->0)(f(x+h)- f(x))/h

Using f(x) = sin 2x, the derivative is:

lim_(h->0)(sin(2*(x+h))- sin 2x)/h

=> lim_(h->0)(sin(2x+2h)- sin 2x)/h

=> lim_(h->0)(2*cos((2x + 2h + 2x)/2)*sin((2x+2h - 2x)/2))/h

=> lim_(h->0)(2*cos (2x+h)*sin h)/h

=> lim_(h->0)2*cos (2x+h)*(sin h)/h

=> lim_(h->0)2*cos (2x+h)*lim_(h->0)(sin h/h)

Use lim_(h->0)(sin h/h) = 1 and substituting h = 0

=> 2*cos 2x

The derivative of sin 2x is 2*cos 2x