Derivative of sin(2x) by first principle
Understand the Problem
The question is asking to find the derivative of the function sin(2x) using the first principle of derivatives. This involves using the limit definition of the derivative, which is the limit of the difference quotient as the increment approaches zero.
Answer
The derivative is \( f'(x) = 2 \cos(2x) \).
Answer for screen readers
The derivative of the function ( \sin(2x) ) is ( f'(x) = 2 \cos(2x) ).
Steps to Solve
- Define the function and the difference quotient
Let ( f(x) = \sin(2x) ). The derivative using the first principle is defined as: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$
- Calculate ( f(x+h) )
Substituting ( x+h ) into the function gives us: $$ f(x+h) = \sin(2(x+h)) = \sin(2x + 2h) $$
- Set up the difference quotient
Now we can write the difference quotient: $$ f'(x) = \lim_{h \to 0} \frac{\sin(2x + 2h) - \sin(2x)}{h} $$
- Use the sine difference identity
Utilizing the sine difference identity ( \sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) ): $$ \sin(2x + 2h) - \sin(2x) = 2 \cos\left(2x + h\right) \sin(h) $$ Thus, substituting this into the difference quotient results in: $$ f'(x) = \lim_{h \to 0} \frac{2 \cos(2x + h) \sin(h)}{h} $$
- Split the limit
We can rewrite this limit as: $$ f'(x) = 2 \lim_{h \to 0} \cos(2x + h) \cdot \lim_{h \to 0} \frac{\sin(h)}{h} $$
- Evaluate the limits
The limit ( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 ). Thus, we find: $$ f'(x) = 2 \cos(2x) \cdot 1 = 2 \cos(2x) $$
The derivative of the function ( \sin(2x) ) is ( f'(x) = 2 \cos(2x) ).
More Information
Using the first principle of derivatives allows us to derive the rate of change of functions like ( \sin(2x) ). The factor of ( 2 ) in the cosine function appears due to the chain rule, combined with our limit definition approach.
Tips
- Forgetting to apply the sine difference identity correctly.
- Not simplifying the difference quotient fully before taking the limit.
- Confusing the limit ( \lim_{h \to 0} \frac{\sin(h)}{h} ) with other limits; it's specifically equal to 1.
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