differentiate sin(2x) cos(2x)
Understand the Problem
The question is asking for the differentiation of the function sin(2x) * cos(2x). To solve this, we will use the product rule of differentiation which states that if you have two functions u and v, the derivative of their product uv is given by u'v + uv'. Here, u = sin(2x) and v = cos(2x). We will differentiate both functions and apply the product rule.
Answer
The derivative of $\sin(2x) \cos(2x)$ is $2\cos(4x)$.
Answer for screen readers
The derivative of the function $\sin(2x) \cos(2x)$ is $2\cos(4x)$.
Steps to Solve
- Define the functions
Let $u = \sin(2x)$ and $v = \cos(2x)$.
- Differentiate u
Now we differentiate $u$. Using the chain rule, we get: $$ u' = \frac{d}{dx}(\sin(2x)) = 2\cos(2x) $$
- Differentiate v
Next, we differentiate $v$ using the chain rule: $$ v' = \frac{d}{dx}(\cos(2x)) = -2\sin(2x) $$
- Apply the product rule
Now, we will apply the product rule. The product rule states: $$ \frac{d}{dx}(uv) = u'v + uv' $$ Substituting $u$, $v$, $u'$, and $v'$ into the product rule gives: $$ \frac{d}{dx}(\sin(2x) \cos(2x)) = (2\cos(2x))(\cos(2x)) + (\sin(2x))(-2\sin(2x)) $$
- Simplify the expression
Now we simplify the resulting expression: $$ = 2\cos^2(2x) - 2\sin^2(2x) $$ This can be rewritten using a trigonometric identity: $$ = 2(\cos^2(2x) - \sin^2(2x)) $$ Which further simplifies to: $$ = 2\cos(4x) $$
The derivative of the function $\sin(2x) \cos(2x)$ is $2\cos(4x)$.
More Information
The result can also be interpreted using double angle identities in trigonometry. The function $\sin(2x) \cos(2x)$ can be rewritten using the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, demonstrating how derivatives often relate to trigonometric identities.
Tips
- Forgetting to apply the chain rule when differentiating functions like $\sin(2x)$ or $\cos(2x)$.
- Not applying the product rule correctly, such as mixing up $u'$ and $v'$ terms or signs.
- Failing to simplify the final result when possible.
