Let's go through each of these derivatives step-by-step. I'll find $\frac{dy}{dx}$ for each given function and simplify where possible.

Problem 1

$y = \frac{1}{2}x + \frac{1}{4} \sin 2x$

To find $\frac{dy}{dx}$:

1. The derivative of $\frac{1}{2}x$ with respect to $x$ is $\frac{1}{2}$.
2. For $\frac{1}{4} \sin 2x$, use the chain rule:
   • The derivative of $\sin 2x$ is $2 \cos 2x$.
   • Therefore, $\frac{d}{dx} \left( \frac{1}{4} \sin 2x \right) = \frac{1}{4} \cdot 2 \cos 2x = \frac{1}{2} \cos 2x$.

So, $\frac{dy}{dx} = \frac{1}{2} + \frac{1}{2} \cos 2x.$

Problem 2

$y = \sin 5x - \frac{1}{3} \sin^3 5x$

To find $\frac{dy}{dx}$:

1. For $\sin 5x$, use the chain rule:
   • The derivative is $5 \cos 5x$.
2. For $\frac{1}{3} \sin^3 5x$, use the chain rule and the power rule:
   • Rewrite as $\frac{1}{3} (\sin 5x)^3$.
   • The derivative is $\frac{1}{3} \cdot 3 (\sin 5x)^2 \cdot 5 \cos 5x = 5 \sin^2 5x \cos 5x$.

So, $\frac{dy}{dx} = 5 \cos 5x - 5 \sin^2 5x \cos 5x.$

This can be simplified by factoring out $5 \cos 5x$: $\frac{dy}{dx} = 5 \cos 5x (1 - \sin^2 5x).$

Using the identity $1 - \sin^2 \theta = \cos^2 \theta$, we get: $\frac{dy}{dx} = 5 \cos 5x \cdot \cos^2 5x = 5 \cos^3 5x.$

Problem 3

$y = 3 \tan 2x + \tan^3 2x$

To find $\frac{dy}{dx}$:

1. For $3 \tan 2x$, use the chain rule:
   • The derivative of $\tan 2x$ is $2 \sec^2 2x$.
   • So, $\frac{d}{dx} (3 \tan 2x) = 3 \cdot 2 \sec^2 2x = 6 \sec^2 2x$.
2. For $\tan^3 2x$, use the chain rule and the power rule:
   • Rewrite as $(\tan 2x)^3$.
   • The derivative is $3 (\tan 2x)^2 \cdot 2 \sec^2 2x = 6 \tan^2 2x \sec^2 2x$.

So, $\frac{dy}{dx} = 6 \sec^2 2x + 6 \tan^2 2x \sec^2 2x.$

Problem 4

$y = -\frac{3}{5} \cot^5 \frac{x}{3} + \cot^3 \frac{x}{3} - 3 \cot \frac{x}{3} - x$

To find $\frac{dy}{dx}$:

1. For $-\frac{3}{5} \cot^5 \frac{x}{3}$, use the chain rule:
   • The derivative is $-\frac{3}{5} \cdot 5 \cot^4 \frac{x}{3} \cdot \left( -\frac{1}{3} \csc^2 \frac{x}{3} \right) = \frac{3}{3} \cot^4 \frac{x}{3} \csc^2 \frac{x}{3} = \cot^4 \frac{x}{3} \csc^2 \frac{x}{3}$.
2. For $\cot^3 \frac{x}{3}$, use the chain rule:
   • The derivative is $3 \cot^2 \frac{x}{3} \cdot \left( -\frac{1}{3} \csc^2 \frac{x}{3} \right) = -\cot^2 \frac{x}{3} \csc^2 \frac{x}{3}$.
3. For $-3 \cot \frac{x}{3}$, the derivative is $-3 \cdot \left( -\frac{1}{3} \csc^2 \frac{x}{3} \right) = \csc^2 \frac{x}{3}$.
4. For $-x$, the derivative is $-1$.

So, $\frac{dy}{dx} = \cot^4 \frac{x}{3} \csc^2 \frac{x}{3} - \cot^2 \frac{x}{3} \csc^2 \frac{x}{3} + \csc^2 \frac{x}{3} - 1.$

Let me know if you need further details or explanations for any of these solutions!

Follow-up Questions

1. How would you differentiate a composite trigonometric function like $\sin^n(f(x))$?
2. What trigonometric identities might help in simplifying derivatives further?
3. Could you apply a similar method to find higher-order derivatives for these functions?
4. How does the chain rule differ from the product and quotient rules in derivatives?
5. How can you apply these derivatives in real-world applications involving periodic motion?

Tip

When dealing with trigonometric functions raised to a power, always apply the chain rule carefully to avoid missing inner derivatives.