To find the derivative of the function $y = \cos^{-1}(2x)$ using implicit differentiation, we will differentiate both sides with respect to $x$.

Step-by-Step Solution:

1. Start with the given function: $y = \cos^{-1}(2x)$

2. Apply the cosine inverse function to both sides: $\cos(y) = 2x$

3. Differentiate both sides implicitly with respect to $x$: [ \frac{d}{dx} [\cos(y)] = \frac{d}{dx} [2x] ]

4. Compute the derivatives:

   • The derivative of $\cos(y)$ with respect to $x$ is $-\sin(y) \frac{dy}{dx}$.
   • The derivative of $2x$ with respect to $x$ is $2$.

   Thus: $-\sin(y) \frac{dy}{dx} = 2$

5. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{2}{\sin(y)}$

6. Express $\sin(y)$ in terms of $x$:

   Since $\cos(y) = 2x$, we use the Pythagorean identity: $\sin^2(y) = 1 - \cos^2(y) = 1 - (2x)^2 = 1 - 4x^2$

   Thus: $\sin(y) = \sqrt{1 - 4x^2}$

   (Note: We choose the positive root because the range of $\cos^{-1}(x)$ is $[0, \pi]$, where $\sin(y) \geq 0$.)

7. Substitute $\sin(y)$ back into the derivative: $\frac{dy}{dx} = -\frac{2}{\sqrt{1 - 4x^2}}$

Final Answer:

The derivative of the function $y = \cos^{-1}(2x)$ is: $\frac{dy}{dx} = -\frac{2}{\sqrt{1 - 4x^2}}.$

Would you like to know more details or have any questions about the steps?

Related Questions:

1. What are the derivatives of other inverse trigonometric functions?
2. How do we use implicit differentiation for functions involving both $x$ and $y$?
3. What is the domain of $y = \cos^{-1}(2x)$?
4. How can we verify the derivative using the chain rule?
5. What happens if we differentiate with respect to another variable?

Tip:

Always check the domain of the original function to ensure the derivative is valid in that domain.