DIFFERENTIATION 08
The Chain Rule
The chain rule differentiates a function built inside another function.
The main idea
Suppose
\[ y=f(u) \]and
\[ u=g(x). \]Then \(y\) depends on \(x\) through the intermediate variable \(u\).
The chain rule is
1. Why a chain appears
If \(x\) changes, then \(u\) changes.
Because \(y\) depends on \(u\), that change in \(u\) causes a change in \(y\).
So the overall rate of change is built from two rates:
\[ \text{change of }y\text{ with respect to }u \]and
\[ \text{change of }u\text{ with respect to }x. \]2. A simple example
Consider
\[ y=(3x+1)^4. \]There is an inner function
\[ u=3x+1 \]and an outer function
\[ y=u^4. \]Differentiate the outer function:
\[ \frac{dy}{du}=4u^3. \]Differentiate the inner function:
\[ \frac{du}{dx}=3. \]Therefore,
\[ \frac{dy}{dx} = 4u^3(3). \]Substitute back \(u=3x+1\):
3. The practical pattern
For
\[ y=[g(x)]^n, \]differentiate the outer power first:
\[ n[g(x)]^{n-1}, \]then multiply by the derivative of the inside:
4. Another example
Let
\[ y=(x^2+5)^3. \]The outer derivative is
\[ 3(x^2+5)^2. \]The inner derivative is
\[ 2x. \]Therefore,
\[ \frac{dy}{dx} = 3(x^2+5)^2(2x). \]So
\[ \boxed{ \frac{dy}{dx} = 6x(x^2+5)^2 } \]5. Chain rule with exponential functions
Consider
\[ y=e^{2x}. \]The outer function is
\[ e^u, \]whose derivative is
\[ e^u. \]The inner function is
\[ u=2x, \]so
\[ \frac{du}{dx}=2. \]Therefore,
6. Chain rule with logarithms
Consider
\[ y=\ln(3x+1). \]For
\[ y=\ln u, \]we have
\[ \frac{dy}{du} = \frac{1}{u}. \]Also,
\[ u=3x+1 \]gives
\[ \frac{du}{dx}=3. \]Therefore,
\[ \frac{dy}{dx} = \frac{1}{3x+1}(3). \]So
\[ \boxed{ \frac{dy}{dx} = \frac{3}{3x+1} } \]7. Chain rule with trigonometric functions
Consider
\[ y=\sin(5x). \]The outer derivative is
\[ \cos(5x), \]and the inner derivative is
\[ 5. \]Therefore,
\[ \boxed{ \frac{dy}{dx} = 5\cos(5x) } \]8. Several layers
The chain rule can be applied repeatedly.
Consider
\[ y=\sin\left((2x+1)^3\right). \]The layers are
\[ x \longrightarrow 2x+1 \longrightarrow (2x+1)^3 \longrightarrow \sin\left((2x+1)^3\right). \]Differentiate from the outside inward:
\[ \frac{dy}{dx} = \cos\left((2x+1)^3\right) \cdot 3(2x+1)^2 \cdot 2. \]Therefore,
\[ \boxed{ \frac{dy}{dx} = 6(2x+1)^2 \cos\left((2x+1)^3\right) } \]9. Why the Leibniz notation is useful
The chain rule can be written
\[ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}. \]The \(du\) terms appear to cancel:
\[ \frac{dy}{\cancel{du}} \frac{\cancel{du}}{dx} = \frac{dy}{dx}. \]This is a useful way to remember the structure.
But the chain rule is the mathematical reason the calculation is valid; it is not justified merely by treating derivatives as ordinary fractions.
10. Function notation form
If
\[ y=f(g(x)), \]then
The derivative of the outer function is evaluated at the inner function, then multiplied by the derivative of the inner function.
11. A useful way to identify the layers
Start from the outside and ask:
What operation is being performed last?
For
\[ (1+x^2)^5, \]the last operation is “raise to the fifth power”.
So the outer function is
\[ u^5 \]and the inner function is
\[ u=1+x^2. \]This makes the chain rule easier to see.
12. Common mistake
For
\[ y=(x^2+1)^4, \]do not write only
\[ 4(x^2+1)^3. \]That differentiates the outer function but ignores the changing inner function.
The correct derivative is
\[ 4(x^2+1)^3(2x). \]Therefore,
\[ \boxed{ \frac{dy}{dx} = 8x(x^2+1)^3 } \]13. The central structure
In symbols,
\[ \boxed{ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} } \]The chain rule is one of the most important ideas in all of calculus.