Chain Rule for Derivatives

Composite Functions

In a composite function, the output of one function becomes the input of another. If $g(x)$ is the inner function and $f(u)$ is the outer function, their composition is $y = f(g(x))$.

The chain rule tells us how to differentiate this nested structure without losing the effect of the inner function.

The Chain Rule Theorem

The inner function stays inside the outer function, so the derivative of the composite function multiplies two rates that act in sequence.

If $y = f(g(x))$, with $g$ differentiable at $x$ and $f$ differentiable at $g(x)$, then the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Formally, if we let $u = g(x)$, then $y = f(u)$. Its derivative is defined as:

This is the chain rule. Work from the outermost layer inward, keep each inner expression in place until its turn, and multiply the derivatives.

Applying the Chain Rule

The following examples show how the same process works with two and three layers. In each one, name the outer function and the inner function before differentiating, because the chain rule multiplies the outer derivative by the inner derivative.

Power Form Functions

Find the first derivative of $y = (x^2 - 4x - 2)^8$.

Solution:

First, we need to break this function down into two parts: an outer function and an inner function.

• Inner function ($u$) is the expression inside the parentheses: $u = x^2 - 4x - 2$.
• Outer function ($y$) is the power operation: $y = u^8$.

Next, we find the derivative of each function:

• Derivative of the inner function: $\frac{du}{dx} = 2x - 4$.
• Derivative of the outer function: $\frac{dy}{du} = 8u^7$.

Now, we can combine them using the chain rule:

Finally, we substitute $u$ back in and simplify the expression:

Trigonometric Functions

Find the first derivative of $y = \cos^4(3 - 4x)$.

Solution:

We can rewrite this function as $y = (\cos(3 - 4x))^4$. This is a case where we need to apply the chain rule more than once because there are three layers of functions.

• Innermost function: $v = 3 - 4x$
• Middle function: $u = \cos(v)$
• Outermost function: $y = u^4$

Then, we differentiate each layer:

• Derivative of the outermost function: $\frac{dy}{du} = 4u^3$
• Derivative of the middle function: $\frac{du}{dv} = -\sin(v)$
• Derivative of the innermost function: $\frac{dv}{dx} = -4$

Combine them all using the chain rule:

Now, substitute $u$ and $v$ back in step-by-step to get the final result:

Exercises

Each problem nests one function inside another, so the chain rule is needed. Identify the outer and the inner function before differentiating.

1. Find the first derivative of $y = (3x^2 - 1)^3$.
2. Find the first derivative of $y = \sin^3(2x)$.

Answer Key

1. Solution for $y = (3x^2 - 1)^3$

   Step 1: Identify the functions

   • Inner function: $u = 3x^2 - 1$
   • Outer function: $y = u^3$

   Step 2: Find the derivative of each

   • $\frac{du}{dx} = 6x$
   • $\frac{dy}{du} = 3u^2$

   Step 3: Combine with the chain rule

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

   $$\frac{dy}{dx} = 3u^2 \cdot 6x = 18x \cdot u^2$$

   Step 4: Don't forget to substitute back

   $$\frac{dy}{dx} = 18x(3x^2 - 1)^2$$

2. Solution for $y = \sin^3(2x)$

   This function is $y = (\sin(2x))^3$. We'll use the multi-layered chain rule.

   Step 1: Identify the functions

   • Innermost function: $v = 2x$
   • Middle function: $u = \sin(v)$
   • Outermost function: $y = u^3$

   Step 2: Find the derivative of each

   • $\frac{dv}{dx} = 2$
   • $\frac{du}{dv} = \cos(v)$
   • $\frac{dy}{du} = 3u^2$

   Step 3: Combine with the chain rule

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

   $$\frac{dy}{dx} = (3u^2) \cdot (\cos(v)) \cdot (2)$$

   Step 4: Substitute back and simplify

   $$\frac{dy}{dx} = 6u^2 \cos(v)$$

   $$\frac{dy}{dx} = 6\sin^2(v) \cos(v)$$

   $$\frac{dy}{dx} = 6\sin^2(2x) \cos(2x)$$