The Product Rule

Consider $f(x)=x^2$ and $g(x)=x^5$. Applying the power rule, we get $f'(x)=2x$ and $g'(x)=5x^4$. Now, $f(x)\cdot g(x)=x^2\cdot x^5=x^7$, so we have $\big(f(x)\cdot g(x)\big)'=7x^6$, again by the power rule. However, $$f'(x)\cdot g'(x)=2x\cdot 5x^4=10x^5\ne 7x^6!$$ In particular, this shows that $\big(f(x)g(x)\big)'\ne f'(x)\cdot g'(x)$.  We differentiate a product by the product rule.


The Product Rule  $$ \frac{d}{dx}\big(f(x)\cdot g(x)\big) = f'(x)\cdot g(x) + f(x)\cdot g'(x)$$

The derivative of a product is not the product of the derivatives.

We write, briefly, that $(fg)'=f'g+fg'$ -- take the derivative of one function and leave the other alone, then add to that the derivative of the other function and leave the first alone.

We can extend the Product Rule to the product of three functions:  $(fgh)'=f'gh+fg'h+fgh'$.

DO:  What do you think $(fghp)'$ is?

Example:   Since the derivative of $x^2$ is $2x$ and the derivative of $e^x$ is $e^x$, the derivative of $x^2e^x$ is $$ (x^2 e^x)' = \frac{d}{dx}(x^2) e^x + x^2 \frac{d}{dx}(e^x) = 2x e^x + x^2 e^x = (x^2+2x)e^x.$$

DO:  Find $\displaystyle\frac{d}{dx}(7x^5e^x)$.