a) determine the derivative of y = sin2x

b) determine the derivative of y = 2SinxCosx

c) shoe that derivatives in parts a) and b) are equal

d) explain why derivatives in parts a) and be should be equal.

a) 2cos2x

b) 2cosx*cosx - 2sinx*sinx = 2cos2x
c) double angle identity

a) To determine the derivative of y = sin(2x), we can use the chain rule. The chain rule states that if we have a composition of functions, f(g(x)), then the derivative is given by the product of the derivative of the outer function (f'(g(x))) and the derivative of the inner function (g'(x)).

In this case, the outer function is sin(u) and the inner function is 2x, where u = 2x.

The derivative of sin(u) with respect to u is cos(u), and the derivative of 2x with respect to x is 2.

Therefore, the derivative of y = sin(2x) is given by:
dy/dx = cos(2x) * 2
= 2cos(2x)

b) To determine the derivative of y = 2sin(x)cos(x), we can again use the product rule. The product rule states that if we have a product of two functions, f(x)g(x), then the derivative is given by the sum of the product of the derivative of the first function (f'(x)) and the second function (g(x)), and the product of the first function (f(x)) and the derivative of the second function (g'(x)).

In this case, the first function is 2sin(x) and the second function is cos(x).

The derivative of 2sin(x) with respect to x is 2cos(x), and the derivative of cos(x) with respect to x is -sin(x).

Therefore, the derivative of y = 2sin(x)cos(x) is given by:
dy/dx = 2cos(x) * cos(x) + 2sin(x) * (-sin(x))
= 2cos^2(x) - 2sin^2(x)

c) To show that the derivatives in parts a) and b) are equal, we can compare the expressions of the derivatives obtained in the previous steps.

In part a), we found that the derivative of y = sin(2x) is 2cos(2x).
In part b), we found that the derivative of y = 2sin(x)cos(x) is 2cos^2(x) - 2sin^2(x).

We can simplify the expression from part b) using the trigonometric identity cos^2(x) - sin^2(x) = cos(2x).

Therefore, the derivative of y = 2sin(x)cos(x) is also 2cos(2x), which is the same as the derivative obtained in part a).

d) The derivatives in parts a) and b) should be equal because sin(2x) and 2sin(x)cos(x) represent the same function. In part a), the function is expressed directly as sin(2x), while in part b), the function is expressed as the product of 2sin(x) and cos(x).

By simplifying the expression in part b) using the trigonometric identity, we can see that both expressions are equivalent. Therefore, their derivatives should be equal. The fact that both derivatives are equal confirms that the functions in parts a) and b) are different ways of expressing the same mathematical relationship.