derivative of sin(2x) cos(2x)
Understand the Problem
The question is asking for the derivative of the function sin(2x) * cos(2x). To solve this, we will need to apply the product rule of differentiation.
Answer
The derivative is $f'(x) = 2\cos(4x)$.
Answer for screen readers
The derivative of the function is given by:
$$ f'(x) = 2\cos(4x) $$
Steps to Solve
- Identify the function and its components
The function we need to differentiate is $f(x) = \sin(2x) \cdot \cos(2x)$. Let's identify the two parts of the product:
Let $u = \sin(2x)$ and $v = \cos(2x)$.
- Apply the product rule
The product rule states that if you have a product of two functions, the derivative is given by:
$$ \frac{d}{dx}(uv) = u'v + uv' $$
Therefore, we will calculate $u'$ and $v'$ first.
- Differentiate u and v
Now, we differentiate $u$ and $v$:
For $u = \sin(2x)$, using the chain rule, we have:
$$ u' = 2\cos(2x) $$
For $v = \cos(2x)$, using the chain rule, we have:
$$ v' = -2\sin(2x) $$
- Substitute u, v, u', and v' into the product rule
Now, we can substitute these values into the product rule formula:
$$ f'(x) = u'v + uv' $$
Replacing $u$, $v$, $u'$, and $v'$ gives:
$$ f'(x) = (2\cos(2x)) \cdot \cos(2x) + \sin(2x) \cdot (-2\sin(2x)) $$
- Simplify the expression
Now let's simplify the expression:
$$ f'(x) = 2\cos^2(2x) - 2\sin^2(2x) $$
This can be factored using the identity $2\cos^2(a) - 2\sin^2(a) = 2\cos(2a)$:
$$ f'(x) = 2(\cos^2(2x) - \sin^2(2x)) = 2\cos(4x) $$
The derivative of the function is given by:
$$ f'(x) = 2\cos(4x) $$
More Information
The derivative we found indicates the rate of change of the function $f(x) = \sin(2x) \cdot \cos(2x)$ at any point $x$. The result $2\cos(4x)$ reflects a transformation of the sine and cosine functions due to their product, highlighting the periodic nature of the result.
Tips
- Forgetting to use the chain rule when differentiating functions involving trigonometric functions with inner functions.
- Neglecting to apply the product rule correctly or mixing up $u'$ and $v'$ during substitution.
