If y=2xsin2x+cos2x, find d²y/dx² if x=2p/3 in exact value
y' = 2sin2x + 4xcos2x - 2sin2x = 4xcos2x
y" = 4cos2x - 8xsin2x
so, just plug in your value for x ...
To find the second derivative of y with respect to x, we need to take the derivative of the first derivative of y.
First, let's find the first derivative of y:
dy/dx = 2(1)sin(2x) + 2xcos(2x) - 2(1)sin(2x) - 2xcos(2x)
dy/dx = 2xcos(2x)
Now, let's find the second derivative of y by taking the derivative of the first derivative:
d²y/dx² = d/dx(2xcos(2x))
To solve this, we will use the product rule, which states that:
d/dx(f(x)*g(x)) = f'(x)*g(x) + f(x)*g'(x)
Using the product rule, we can find the derivative of 2x with respect to x, and the derivative of cos(2x) with respect to x.
Let's start with the derivative of 2x:
d/dx(2x) = 2
Now, let's find the derivative of cos(2x):
d/dx(cos(2x)) = -sin(2x) * 2
So, applying the product rule, we have:
d²y/dx² = (2)(cos(2x)) + (2x)(-sin(2x) * 2)
Simplifying this expression, we get:
d²y/dx² = 2cos(2x) - 4xsin(2x)
Finally, to find the value of d²y/dx² when x = 2p/3, we substitute this value into the expression:
d²y/dx² = 2cos(2(2p/3)) - 4(2p/3)sin(2(2p/3))
Simplifying further, we get the exact value of d²y/dx² at x = 2p/3.