How is 2cos(4x) the derivative of f(x)=sin(2x)cos(2x)?
If you take the derivative of sin2xcos2x, you get a form of cos^2 2x + sin^2 2x, then you use the double angle formulas to get cos4x
You can also start with identity
f(x) = sin 2x cos 2x = (1/2) sin 4x
Then take the derivative with respect to x. Using the chain rule with u(x) = 4x,
df/dx = df/du du/dx
you get
df/dx = (1/2) cos (4x) *4 = 2 cos (4x)
drwls: thanks for your help. would you mind explaining how you used that identity to set up the new problem? i get the chain rule but i had no idea that sin2xcos2x was equal to 1/2sin4x. thanks again.
To find the derivative of the function f(x) = sin(2x)cos(2x), we can use the product rule. The product rule states that if we have a product of two functions, u(x) and v(x), then the derivative of their product is given by:
(fg)'(x) = u'(x)v(x) + u(x)v'(x)
In this case, u(x) = sin(2x) and v(x) = cos(2x). Let's find the derivatives of u(x) and v(x):
u'(x) = derivative of sin(2x) with respect to x
= 2cos(2x) (using the chain rule)
v'(x) = derivative of cos(2x) with respect to x
= -2sin(2x) (using the chain rule)
Now, we can apply the formula for the product rule and find the derivative of f(x):
f'(x) = u'(x)v(x) + u(x)v'(x)
= (2cos(2x))(cos(2x)) + (sin(2x))(-2sin(2x))
= 2cos^2(2x) - 2sin^2(2x)
= 2(cos^2(2x) - sin^2(2x))
Now, we can use the trigonometric identity cos(2θ) = 1 - 2sin^2(θ) to simplify the expression:
f'(x) = 2(1 - 2sin^2(2x) - sin^2(2x))
= 2(1 - 3sin^2(2x))
Finally, using another trigonometric identity sin^2(θ) = 1 - cos^2(θ), we can rewrite the derivative in terms of cos(2x):
f'(x) = 2(1 - 3(1 - cos^2(2x)))
= 2(1 - 3 + 3cos^2(2x))
= 2(3cos^2(2x) - 2)
= 6cos^2(2x) - 4
Therefore, the derivative of f(x) = sin(2x)cos(2x) is 6cos^2(2x) - 4.