Consider the function
y(x)=cos(2x), Find the derivative,
dy/dx and evaluate it at the x-coordinates in the table below. x=-1.51

To find the derivative of the function \( y(x) = \cos(2x) \), we'll use basic differentiation rules. The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \), and we must apply the chain rule because the argument of the cosine function is \( 2x \), not just \( x \).

### Step 1: Differentiate \( y(x) = \cos(2x) \)
\[
\frac{dy}{dx} = \frac{d}{dx}[\cos(2x)]
\]
Using the chain rule:
\[
\frac{dy}{dx} = -\sin(2x) \cdot \frac{d}{dx}[2x]
\]
\[
\frac{dy}{dx} = -\sin(2x) \cdot 2
\]
\[
\frac{dy}{dx} = -2\sin(2x)
\]

### Step 2: Evaluate the derivative at \( x = -1.51 \)
Substitute \( x = -1.51 \) into the derivative:
\[
\frac{dy}{dx}\bigg|_{x=-1.51} = -2\sin(2(-1.51))
\]
\[
\frac{dy}{dx}\bigg|_{x=-1.51} = -2\sin(-3.02)
\]
Using the fact that \( \sin(-\theta) = -\sin(\theta) \):
\[
\frac{dy}{dx}\bigg|_{x=-1.51} = -2(-\sin(3.02))
\]
\[
\frac{dy}{dx}\bigg|_{x=-1.51} = 2\sin(3.02)
\]

Finally, calculate the value of \( \sin(3.02) \):
\[
\sin(3.02) \approx 0.1259
\]
\[
\frac{dy}{dx}\bigg|_{x=-1.51} \approx 2 \times 0.1259 \approx 0.2518
\]

### Final Answer:
\[
\frac{dy}{dx}\bigg|_{x=-1.51} \approx 0.2518
\]

Would you like more details or have any questions?

Here are five related questions:
1. How does the chain rule apply in differentiation?
2. What is the general formula for differentiating trigonometric functions?
3. How can we verify the derivative of a trigonometric function?
4. What is the physical interpretation of the derivative of a function?
5. How do you compute the sine of an angle without a calculator?

**Tip:** When working with trigonometric functions, remember that their derivatives often involve using the chain rule when the argument of the trigonometric function is more complex than a simple \( x \).

Learn how to find the derivative of y(x) = cos(2x) using the chain rule in differentiation. Evaluate the derivative at x = -1.51 and understand the application of trigonometric function derivatives. Suitable for advanced high school students and anyone studying calculus.

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