DIFFERENTIATION 06
The Product Rule
When two changing functions are multiplied, the change in their product depends on the change in both functions.
The main idea
Suppose
\[ y=f(x)g(x). \]The derivative is not simply \(f'(x)g'(x)\).
Instead,
This is the product rule.
1. Why do we need a special rule?
If both \(f(x)\) and \(g(x)\) change when \(x\) changes, then their product changes because of both changes.
For a small change in \(x\),
\[ f \rightarrow f+\Delta f \]and
\[ g \rightarrow g+\Delta g. \]The new product is
\[ (f+\Delta f)(g+\Delta g). \]2. Expand the new product
Expanding gives
\[ fg+g\Delta f+f\Delta g+\Delta f\Delta g. \]The original product was
\[ fg. \]Therefore the change in the product is
This shows why both changing factors contribute to the total change.
3. Divide by the change in \(x\)
Divide by \(\Delta x\):
\[ \frac{\Delta(fg)}{\Delta x} = g\frac{\Delta f}{\Delta x} + f\frac{\Delta g}{\Delta x} + \frac{\Delta f\Delta g}{\Delta x}. \]Rewrite the last term as
\[ \frac{\Delta f\Delta g}{\Delta x} = \Delta f \frac{\Delta g}{\Delta x}. \]4. Take the limit
As
\[ \Delta x\to0, \]for differentiable functions,
\[ \Delta f\to0. \]Therefore,
\[ \Delta f \frac{\Delta g}{\Delta x} \to0. \]The remaining terms give
\[ \boxed{ (fg)' = f'g+fg' } \]which is the product rule.
5. A simple way to remember it
For
\[ y=fg, \]differentiate one factor at a time:
First differentiate \(f\) and leave \(g\) unchanged. Then leave \(f\) unchanged and differentiate \(g\). Finally add the two results.
6. Example
Let
\[ y=x^2(x+3). \]Take
\[ f(x)=x^2 \]and
\[ g(x)=x+3. \]Then
\[ f'(x)=2x \]and
\[ g'(x)=1. \]Using the product rule,
\[ \frac{dy}{dx} = (2x)(x+3) + x^2(1). \]Therefore,
\[ \frac{dy}{dx} = 2x^2+6x+x^2. \]So
7. Check by expanding first
The original function is
\[ y=x^2(x+3). \]Expanding,
\[ y=x^3+3x^2. \]Differentiate:
\[ \frac{dy}{dx} = 3x^2+6x. \]This agrees with the product rule.
8. Example where expanding is not convenient
Consider
\[ y=x^2\sin x. \]Let
\[ f(x)=x^2, \qquad g(x)=\sin x. \]Then
\[ f'(x)=2x, \qquad g'(x)=\cos x. \]Therefore,
The product rule is useful here because the product cannot be simplified by ordinary algebra before differentiating.
9. Example with an exponential function
Suppose
\[ y=xe^x. \]Then
\[ f(x)=x, \qquad g(x)=e^x. \]Their derivatives are
\[ f'(x)=1, \qquad g'(x)=e^x. \]Therefore,
\[ \frac{dy}{dx} = (1)e^x+x(e^x). \]Hence,
10. Three factors
The same idea extends to three functions.
If
\[ y=fgh, \]differentiate one factor at a time:
Each term contains exactly one differentiated factor.
11. A common mistake
Do not write
\[ (fg)'=f'g'. \]For example,
\[ x^2x^3=x^5. \]The correct derivative is
\[ 5x^4. \]But multiplying the separate derivatives would give
\[ (2x)(3x^2)=6x^3, \]which is different.
12. The structure to remember
In symbols,
\[ \boxed{ f'g+fg' } \]The product changes because either factor can contribute to the change.