1. Identify inner ($u = x^2$) and outer ($y = \sin(u)$) functions. 2. Find $\frac{dy}{du} = \cos(u)$. 3. Find $\frac{du}{dx} = 2x$. 4. Multiply: $\frac{dy}{dx} = \cos(x^2) \cdot 2x = 2x\cos(x^2)$

1. Inner: $u = \sqrt{x}$, Outer: $f(u) = e^u$. 2. $f'(u) = e^u$. 3. $u'(x) = \frac{1}{2\sqrt{x}}$. 4. $f'(x) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}}$

1. Identify inner function $u = 3x^2 + 1$ and outer function $y = u^5$. 2. Find $\frac{du}{dx} = 6x$. 3. Find $\frac{dy}{du} = 5u^4$. 4. Multiply: $\frac{dy}{dx} = 5(3x^2 + 1)^4 cdot 6x = 30x(3x^2 + 1)^4$.

1. Rewrite as $y = (\cos(x))^2$. 2. Inner: $u = \cos(x)$, Outer: $y = u^2$. 3. $\frac{du}{dx} = -\sin(x)$. 4. $\frac{dy}{du} = 2u$. 5. $\frac{dy}{dx} = 2\cos(x) cdot -\sin(x) = -2\sin(x)\cos(x)$

1. Inner: $u = \sin(x)$, Outer: $y = \sqrt{u}$. 2. $\frac{du}{dx} = \cos(x)$. 3. $\frac{dy}{du} = \frac{1}{2\sqrt{u}}$. 4. $\frac{dy}{dx} = \frac{1}{2\sqrt{\sin(x)}} \cdot \cos(x) = \frac{\cos(x)}{2\sqrt{\sin(x)}}$

1. Identify inner function $u = 5x^2 + 2x$ and outer function $y = e^u$. 2. Find $\frac{du}{dx} = 10x + 2$. 3. Find $\frac{dy}{du} = e^u$. 4. Multiply: $\frac{dy}{dx} = e^{5x^2 + 2x} cdot (10x + 2) = (10x + 2)e^{5x^2 + 2x}$.

1. Inner: $u = x^3 + 1$, Outer: $y = \ln(u)$. 2. $\frac{du}{dx} = 3x^2$. 3. $\frac{dy}{du} = \frac{1}{u}$. 4. $\frac{dy}{dx} = \frac{1}{x^3 + 1} cdot 3x^2 = \frac{3x^2}{x^3 + 1}$.

1. Inner: $u = x^2 + 3x - 1$, Outer: $y = u^2$. 2. $\frac{du}{dx} = 2x + 3$. 3. $\frac{dy}{du} = 2u$. 4. $\frac{dy}{dx} = 2(x^2 + 3x - 1) cdot (2x + 3)$.

1. Inner: $u = 5x^3 + 2x^2 + 6$, Outer: $y = 4u^2$. 2. $\frac{du}{dx} = 15x^2 + 4x$. 3. $\frac{dy}{du} = 8u$. 4. $\frac{dy}{dx} = 8(5x^3 + 2x^2 + 6) cdot (15x^2 + 4x)$.

1. Inner: $u = 7x^2$, Outer: $y = \sqrt{u}$. 2. $\frac{du}{dx} = 14x$. 3. $\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2}(7x^2)^{-\frac{1}{2}}$. 4. $\frac{dy}{dx} = \frac{1}{2}(7x^2)^{-\frac{1}{2}} cdot (14x)$.

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F(x) = f(g(x))$ is differentiable at $x$ and $F'(x)$ is given by $F'(x) = f'(g(x)) \cdot g'(x)$.

It provides a formal justification for the method of differentiating composite functions, ensuring that the derivative is calculated correctly by accounting for the rates of change of both the inner and outer functions.

The theorem requires that both the inner function $g(x)$ and the outer function $f(x)$ be differentiable at their respective points for the composite function to be differentiable.

$g$ must be differentiable at $x$, and $f$ must be differentiable at $g(x)$.

It allows us to relate the rates of change of different variables in an equation by differentiating with respect to time and applying the Chain Rule to each term, ensuring that all rates of change are properly accounted for.

If $f$ and $g$ are inverse functions, then $f(g(x)) = x$. Differentiating both sides using the Chain Rule gives $f'(g(x)) \cdot g'(x) = 1$, which can be used to find the derivative of the inverse function.

When dealing with complex functions involving multiple layers of composition, such as $y = \sin(e^{x^2})$, the Chain Rule Theorem is essential for systematically differentiating each layer and obtaining the correct derivative.

It ensures the correct derivative by accounting for the rates of change of both the inner and outer functions and multiplying them together, providing a complete picture of how the composite function changes with respect to its input.

The Chain Rule Theorem relies on the concept of local linearity, where differentiable functions can be approximated by linear functions at a point, allowing us to break down the differentiation of composite functions into smaller, linear approximations.

When differentiating a power function where the base is a function of $x$, such as $y = [f(x)]^n$, the Chain Rule Theorem is used to account for the derivative of the base, resulting in $y' = n[f(x)]^{n-1} \cdot f'(x)$.

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

$\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$

$\frac{dy}{dx} = n \cdot u^{n-1} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = e^{g(x)} \cdot g'(x)$

$\frac{dy}{dx} = \cos(g(x)) \cdot g'(x)$

$\frac{dy}{dx} = -\sin(g(x)) \cdot g'(x)$

$\frac{dy}{dx} = \frac{1}{g(x)} \cdot g'(x) = \frac{g'(x)}{g(x)}$

$y' = n[f(x)]^{n-1} cdot f'(x)$

$\frac{dy}{dx} = \sec^2(g(x)) \cdot g'(x)$

$\frac{dy}{dx} = \frac{g'(x)}{2\sqrt{g(x)}}$