Chain Rule for Derivatives

Sahabat Smaridasa - Mathematical Concepts

Previously, we discussed "Derivative of Algebraic Functions" and the basic derivative rules, including the chain rule. In this article, we will explore the Chain Rule for Derivatives in more detail. The chain rule is used to differentiate composite functions (functions that depend on other functions).

Explanation of the Chain Rule

Suppose we have a composite function \(y = f[g(x)]\). To find its derivative, we use the chain rule.

Let \(z = g(x)\). Then the function becomes \(y = f(z)\).

For \(z = g(x)\), we have \(\frac{dz}{dx} = g'(x)\).

For \(y = f(z)\), we have \(\frac{dy}{dz} = f'(z) = f'[g(x)]\).

Thus, the derivative of \(y = f[g(x)]\) using the chain rule is:

\[ y' = \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = f'[g(x)] \cdot g'(x) \]

For the special case \(y = [g(x)]^n\), let \(z = g(x)\). Then \(y = z^n\). We have:

\[ \frac{dy}{dz} = n z^{n-1} = n [g(x)]^{n-1}, \quad \frac{dz}{dx} = g'(x) \]

Therefore:

\[ y' = \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = n [g(x)]^{n-1} \cdot g'(x) \]

📘 Example 1 : Power of a Function

Find the derivative of \(y = (x^3 - 2x + 2)^{2015}\) and determine \(f'(1)\).

Solution:

Let \(z = x^3 - 2x + 2\). Then \(\frac{dz}{dx} = 3x^2 - 2\).

The function becomes \(y = z^{2015}\), so \(\frac{dy}{dz} = 2015 z^{2014} = 2015 (x^3 - 2x + 2)^{2014}\).

Using the chain rule:

\[ y' = \frac{dy}{dz} \cdot \frac{dz}{dx} = 2015 (x^3 - 2x + 2)^{2014} \cdot (3x^2 - 2) \]

Thus, \(f'(x) = 2015 (x^3 - 2x + 2)^{2014} (3x^2 - 2)\).

Now evaluate \(f'(1)\):

\[ f'(1) = 2015 (1^3 - 2 \cdot 1 + 2)^{2014} \cdot (3 \cdot 1^2 - 2) = 2015 (1)^{2014} \cdot 1 = 2015 \]

Therefore, \(f'(1) = 2015\).

📘 Example 2 : Chain Rule with Product of Functions

Given \(g(2x - 3) = 2x^2 \cdot f(x^2 - 1)\) with \(f(3) = -2\) and \(f'(3) = 1\), find \(g'(1)\).

Solution:

Differentiate both sides of the equation with respect to \(x\).

Left side: \(y = g(2x - 3)\) ⇒ \(y' = g'(2x - 3) \cdot 2 = 2g'(2x - 3)\).

Right side: \(y = 2x^2 \cdot f(x^2 - 1)\). Let \(U = 2x^2\) and \(V = f(x^2 - 1)\).

\(U' = 4x\). For \(V'\), use the chain rule: \(V' = f'(x^2 - 1) \cdot 2x = 2x f'(x^2 - 1)\).

Using the product rule: \(y' = U'V + UV' = 4x \cdot f(x^2 - 1) + 2x^2 \cdot 2x f'(x^2 - 1) = 4x f(x^2 - 1) + 4x^3 f'(x^2 - 1)\).

Equating both sides: \(2g'(2x - 3) = 4x f(x^2 - 1) + 4x^3 f'(x^2 - 1)\).

We want \(g'(1)\), so set \(2x - 3 = 1\) ⇒ \(x = 2\).

Substitute \(x = 2\):

\[ 2g'(1) = 4(2) f(2^2 - 1) + 4(2^3) f'(2^2 - 1) = 8 f(3) + 32 f'(3) \] \[ 2g'(1) = 8(-2) + 32(1) = -16 + 32 = 16 \] \[ g'(1) = \frac{16}{2} = 8 \]

Thus, \(g'(1) = 8\).

📘 Example 3 : Nested Composite Functions

Given \(f(1) = 1\) and \(f'(1) = 2\), find \(g'(1)\) for \(g(x) = f(f(f(f(f(f(x))))))\) (six compositions).

Solution:

Let us define the nested functions step by step:

• \(z = f(x)\) ⇒ \(\frac{dz}{dx} = f'(x)\)
• \(m = f(f(x)) = f(z)\) ⇒ \(\frac{dm}{dz} = f'(z) = f'(f(x))\)
• \(n = f(f(f(x))) = f(m)\) ⇒ \(\frac{dn}{dm} = f'(m) = f'(f(f(x)))\)
• \(p = f(f(f(f(x)))) = f(n)\) ⇒ \(\frac{dp}{dn} = f'(n) = f'(f(f(f(x))))\)
• \(q = f(f(f(f(f(x))))) = f(p)\) ⇒ \(\frac{dq}{dp} = f'(p) = f'(f(f(f(f(x)))))\)
• \(g(x) = f(f(f(f(f(f(x)))))) = f(q)\) ⇒ \(\frac{dg}{dq} = f'(q) = f'(f(f(f(f(f(x))))))\)

By the chain rule, \(g'(x) = \frac{dg}{dq} \cdot \frac{dq}{dp} \cdot \frac{dp}{dn} \cdot \frac{dn}{dm} \cdot \frac{dm}{dz} \cdot \frac{dz}{dx}\).

At \(x = 1\), we have:

• \(f(1) = 1\) ⇒ \(f'(1) = 2\)
• \(f(f(1)) = f(1) = 1\) ⇒ \(f'(f(1)) = f'(1) = 2\)
• Similarly, each nested composition gives 1, and each derivative gives 2.

Thus, \(g'(1) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^6 = 64\).

Therefore, \(g'(1) = 64\).

Note: The chain rule can be applied to all types of functions, including algebraic functions, trigonometric functions, exponential functions, logarithmic functions, and their combinations. It is one of the most important differentiation rules.