It allows us to differentiate composite functions, which are functions within functions, by breaking down the differentiation process into smaller, manageable steps.

Differentiate the outermost function first, treating the inner function as a single variable, then multiply by the derivative of the inner function.

It's often used in conjunction with other rules like the Product Rule and Quotient Rule when dealing with more complex functions.

You will get an incorrect derivative, as you've only differentiated the outer function and not accounted for the inner function's rate of change.

The inner function is the one 'inside' another function, while the outer function is the one that 'encloses' the inner function.

The Chain Rule is specifically designed to differentiate composite functions, so understanding their structure is essential for applying the rule correctly.

Apply the Chain Rule iteratively, starting with the outermost function and working your way inwards, multiplying by the derivative of each inner function at each step.

Forgetting to apply the chain rule to every layer of the function. Don't forget the derivative of the innermost function!

The Chain Rule is a fundamental tool in implicit differentiation, where we differentiate equations that are not explicitly solved for one variable in terms of the other.

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.

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

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

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

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

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

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

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

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

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

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

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. $\frac{dy}{dx} = \frac{1}{2}(7x^2)^{-\frac{1}{2}} cdot (14x)$.