# Finding Derivatives Made Easy! Derivative of A Chain of Functions

The chain rule is essential in calculus for differentiating compositions of functions. It states that the derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. This rule simplifies the differentiation of complex nested functions, playing a crucial role in advanced mathematical, scientific, and engineering applications.

**Definition:**

The chain rule is a method in calculus used to find the derivative of a composite function. It’s like a two-step process: first, you take the derivative of the outer function, and then you multiply it by the derivative of the inner function. This helps you understand how changes in one variable affect a chain of functions. For function \( f(x) \) and \( g(x) \) , to find the derivative of \( f(g(x)) \), we have:

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

Let’s consider this example: \( \text{Find the derivative of } (x + 1)^3. \)

- Define the inner and outer functions and their derivatives

\( g(x) = x + 1, \ g'(x) = 1 \)

\( f(u) = u^3, \ f'(u) = 3u^2 \)

- Apply the chain rule

\( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) = 3(x + 1)^2 \cdot 1 \)

**Chain rule for more functions:**

Differentiating a composition of four functions using the chain rule involves taking the derivative of each function sequentially and multiplying them together. It’s like unwrapping nested functions layer by layer, applying derivatives at each step to reveal the rate of change of the entire composition.

\( f = f(u), \ u = u(v), \ v = v(w), \ w = w(x) \)

\( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx} \)

This represents the extended chain rule formula for the case where you have a composition of four functions, allowing you to differentiate \( f(u(v(w(x)))) \) with respect to \( x \).

Here is an example for this formula: \( \text{Find the derivative of } \sin(\cos(x^3)). \)

1.**Define functions**

\( v(x) = x^3 \)

\( u(v) = \cos v \)

\( f(u) = \sin u \)

2. **Define derivatives**

\( v'(x) = 3x^2 \)

\( u'(v) = -\sin v \)

\( f'(u) = \cos u \)

3.** Apply chain rule**

\( \frac{df}{dx} = f'(u) \cdot u'(v) \cdot v'(x) \)

4. **Simplify**

\( = \cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2 \)

This **next example** includes radical and logarithm derivatives, which will be explained later. But if you already know how the derivative of those functions work, then this problem is for you to better understand the chain rule:

\( \text{Find the derivative of } \sqrt{\ln(e^{x^2})}. \)

1.**Define functions**

\( w(x) = x^2 \)

\( v(w) = e^w \)

\( u(v) = \ln(v) \)

\( f(u) = \sqrt{u} \)

2. **Define derivatives**

\( w'(x) = 2x \)

\( v'(w) = e^w \)

\( u'(v) = \frac{1}{v} \)

\( f'(u) = \frac{1}{2\sqrt{u}} \)

3. **Apply chain rule**

\( \frac{df}{dx} = f'(u) \cdot u'(v) \cdot v'(w) \cdot w'(x) \)

4. **Simplify**

\( = \frac{1}{2\sqrt{\ln(e^{x^2})}} \cdot \frac{1}{e^{x^2}} \cdot e^{x^2} \cdot 2x \)

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