Let's solve each of the given problems step by step using the chain rule for derivatives.

Problem 1:

$y = \left[x + 3 + (x - 5)^3 \right]^{-\frac{1}{2}}$

Step 1: Identify the outer function and inner function

The outer function is the power function $u^{-\frac{1}{2}}$ where $u = x + 3 + (x - 5)^3$, and the inner function is the expression inside the parentheses.

Step 2: Differentiate using the chain rule

The chain rule states that if $y = f(u)$ and $u = g(x)$, then:

$\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$

1. Outer Function Derivative: $\frac{d}{du} \left(u^{-\frac{1}{2}}\right) = -\frac{1}{2} u^{-\frac{3}{2}}$
2. Inner Function Derivative: $u = x + 3 + (x - 5)^3$ Differentiate each term: $\frac{d}{dx}\left(x + 3 + (x - 5)^3\right) = 1 + 3(x - 5)^2$

Step 3: Apply the chain rule

Now, applying the chain rule:

$\frac{dy}{dx} = -\frac{1}{2} \left( x + 3 + (x - 5)^3 \right)^{-\frac{3}{2}} \cdot \left( 1 + 3(x - 5)^2 \right)$

Problem 2:

$y = \left(\frac{x^2}{10} - 3x\right)^{\frac{1}{3}}$

Step 1: Identify the outer and inner functions

The outer function is $u^{\frac{1}{3}}$, where $u = \frac{x^2}{10} - 3x$.

Step 2: Differentiate using the chain rule

1. Outer Function Derivative: $\frac{d}{du} \left( u^{\frac{1}{3}} \right) = \frac{1}{3} u^{-\frac{2}{3}}$
2. Inner Function Derivative: $u = \frac{x^2}{10} - 3x$ Differentiate: $\frac{d}{dx} \left( \frac{x^2}{10} - 3x \right) = \frac{2x}{10} - 3 = \frac{x}{5} - 3$

Step 3: Apply the chain rule

$\frac{dy}{dx} = \frac{1}{3} \left( \frac{x^2}{10} - 3x \right)^{-\frac{2}{3}} \cdot \left( \frac{x}{5} - 3 \right)$

Problem 3:

$y = \frac{3}{\sqrt{16 + x^4}}$

Step 1: Identify the outer and inner functions

The outer function is $\frac{3}{u^{\frac{1}{2}}}$, where $u = 16 + x^4$.

Step 2: Differentiate using the chain rule

1. Outer Function Derivative: $\frac{d}{du} \left( \frac{3}{u^{\frac{1}{2}}} \right) = -\frac{3}{2} u^{-\frac{3}{2}}$
2. Inner Function Derivative: $u = 16 + x^4$ Differentiate: $\frac{d}{dx} \left( 16 + x^4 \right) = 4x^3$

Step 3: Apply the chain rule

$\frac{dy}{dx} = -\frac{3}{2} \left( 16 + x^4 \right)^{-\frac{3}{2}} \cdot 4x^3$ Simplifying: $\frac{dy}{dx} = -\frac{6x^3}{\left( 16 + x^4 \right)^{\frac{3}{2}}}$

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Summary of Derivatives:

1. $\frac{dy}{dx} = -\frac{1}{2} \left(x + 3 + (x - 5)^3 \right)^{-\frac{3}{2}} \left( 1 + 3(x - 5)^2 \right)$
2. $\frac{dy}{dx} = \frac{1}{3} \left( \frac{x^2}{10} - 3x \right)^{-\frac{2}{3}} \left( \frac{x}{5} - 3 \right)$
3. $\frac{dy}{dx} = -\frac{6x^3}{\left( 16 + x^4 \right)^{\frac{3}{2}}}$

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Do you want further explanation on any of the steps, or is there a specific part you want more details on?

Here are some related questions you can explore next:

1. What happens when the outer function has more complex powers, like $u^n$ where $n$ is not a simple fraction?
2. Can the chain rule be used with multiple layers of functions, and how does it work in those cases?
3. How do you differentiate functions that have compositions of trigonometric or exponential functions?
4. How would you apply the chain rule to higher-order derivatives?
5. How do you use the chain rule in multivariable calculus, like partial derivatives?

Tip: When applying the chain rule, always carefully identify the inner and outer functions. Break the problem down into parts for easier differentiation!