The derivative of sin(x^2)cos(2x) is given by the product rule, which states that the derivative of the product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. Applying this rule, we get:
sin(x^2)cos(2x)' = (sin(x^2))'cos(2x) + sin(x^2)(cos(2x))' = 2xcos(x^2)cos(2x) - sin(x^2)sin(2x)
To simplify the derivative of sin(x^2)cos(2x), we can use the double angle formula for cosine: cos(2x) = 2cos^2(x) - 1. Substituting this into the derivative, we get:
sin(x^2)cos(2x)' = 2xcos(x^2)(2cos^2(x) - 1) - sin(x^2)sin(2x) = 4xcos^3(x^2) - 2xcos(x^2) - sin(x^2)sin(2x)
Yes, you can use the chain rule to find the derivative of sin(x^2)cos(2x). The chain rule states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Applying the chain rule to this function, we get:
sin(x^2)cos(2x)' = (sin(u))'cos(2x) + sin(u)(cos(2x))' = cos(x^2)(2x)cos(2x) + sin(x^2)(-sin(2x)2) = 2xcos(x^2)cos(2x) - 2sin(x^2)sin(2x)
To find the second derivative of sin(x^2)cos(2x), we simply take the derivative of the first derivative. Using the simplified form from question 2, we get:
sin(x^2)cos(2x)'' = (4xcos^3(x^2) - 2xcos(x^2) - sin(x^2)sin(2x))' = 8cos^4(x^2) - 8xcos^2(x^2) + 2sin^2(x^2)cos(2x) - 2xcos(2x) + 2sin(x^2)sin(2x)
To find the critical points of sin(x^2)cos(2x), we need to find the values of x where the derivative is equal to 0 or undefined. From the simplified form in question 2, we can see that the derivative is equal to 0 when cos(x^2) = 0 or sin(x^2) = 0. These occur at the points where x = (2n + 1)π/2, where n is any integer. The derivative is also undefined at x = nπ, where n is any integer. Therefore, the critical points for this function are (2n + 1)π/2 and nπ, where n is any integer.