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# The Product Rule

Consider the product of two simple functions, say $$f(x)=(x^2+1)(x^3-3x)$$. An obvious guess for the derivative of $$f$$ is the product of the derivatives of the constituent functions: $$(2x)(3x^2-3)=6x^3-6x$$.

Is this correct? We can easily check, by rewriting $$f$$ and doing the calculation in a way that is known to work. First, $$f(x)=x^5-3x^3+x^3-3x=x^5-2x^3-3x$$, and then $$f'(x)=5x^4-6x^2-3$$. Not even close! What went "wrong''? Well, nothing really, except the guess was wrong.

So the derivative of $$f(x)g(x)$$ is NOT as simple as $$f'(x)g'(x)$$. Surely there is some rule for such a situation? There is, and it is instructive to "discover'' it by trying to do the general calculation even without knowing the answer in advance.

\eqalign{ {d\over dx}(&f(x)g(x)) = \lim_{\Delta x \to0} {f(x+\Delta x)g(x+\Delta x) - f(x)g(x)\over \Delta x}\cr& =\lim_{\Delta x \to0} {f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x) + f(x+\Delta x)g(x)- f(x)g(x)\over \Delta x}\cr & =\lim_{\Delta x \to0} {f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)\over \Delta x} + \lim_{\Delta x \to0} {f(x+\Delta x)g(x)- f(x)g(x)\over \Delta x}\cr & =\lim_{\Delta x \to0} f(x+\Delta x){ g(x+\Delta x)-g(x)\over \Delta x} + \lim_{\Delta x \to0} {f(x+\Delta x)- f(x)\over \Delta x}g(x)\cr & =f(x)g'(x) + f'(x)g(x)\cr }

A couple of items here need discussion. First, we used a standard trick, "add and subtract the same thing'', to transform what we had into a more useful form. After some rewriting, we realize that we have two limits that produce $$f'(x)$$ and $$g'(x)$$. Of course, $$f'(x)$$ and $$g'(x)$$ must actually exist for this to make sense. We also replaced $$\lim_{\Delta x\to0}f(x+\Delta x)$$ with $$f(x)$$---why is this justified?

What we really need to know here is that $$\lim_{\Delta x\to 0}f(x+\Delta x)=f(x)$$, or in the language of section 2.5, that $$f$$ is continuous at $$x$$. We already know that $$f'(x)$$ exists (or the whole approach, writing the derivative of $$fg$$ in terms of $$f'$$ and $$g'$$, doesn't make sense). This turns out to imply that $$f$$ is continuous as well. Here's why:

\eqalign{ \lim_{\Delta x\to 0} f(x+\Delta x) &= \lim_{\Delta x\to 0} (f(x+\Delta x) -f(x) + f(x))\cr& = \lim_{\Delta x\to 0} {f(x+\Delta x) -f(x)\over \Delta x}\Delta x + \lim_{\Delta x\to 0} f(x)\cr& =f'(x)\cdot 0 + f(x) = f(x)\cr }

To summarize: the product rule says that

${d\over dx}(f(x)g(x)) = f(x)g'(x) + f'(x)g(x).$

Returning to the example we started with, let

$f(x)=(x^2+1)(x^3-3x).$

Then

$f'(x)=(x^2+1)(3x^2-3)+(2x)(x^3-3x)=3x^4-3x^2+3x^2-3+2x^4-6x^2= 5x^4-6x^2-3,$

as before. In this case it is probably simpler to multiply $$f(x)$$ out first, then compute the derivative; here's an example for which we really need the product rule.

Example 3.3.1
Compute the derivative of $$f(x)=x^2\sqrt{625-x^2}$$.
SOLUTION
We have already computed
${d\over dx}\sqrt{625-x^2}={-x\over\sqrt{625-x^2}}.$
Now
$f'(x)=x^2{-x\over\sqrt{625-x^2}}+2x\sqrt{625-x^2}= {-x^3+2x(625-x^2)\over \sqrt{625-x^2}}= {-3x^3+1250x\over \sqrt{625-x^2}}.$

### Exercises 3.3

In 1--4, find the derivatives of the functions using the product rule.

Ex 3.3.1 $$x^3(x^3-5x+10)$$ (answer)

Ex 3.3.2 $$(x^2+5x-3)(x^5-6x^3+3x^2-7x+1)$$ (answer)

Ex 3.3.3 $$\sqrt{x}\sqrt{625-x^2}$$ (answer)

Ex 3.3.4 $$\displaystyle {\sqrt{625-x^2}\over x^{20}}$$ (answer)

Ex 3.3.5 Use the product rule to compute the derivative of $$f(x)=(2x-3)^2$$. Sketch the function. Find an equation of the tangent line to the curve at $$x=2$$. Sketch the tangent line at $$x=2$$. (answer)

Ex 3.3.6 Suppose that $$f$$, $$g$$, and $$h$$ are differentiable functions. Show that $$(fgh)'(x) = f'(x) g(x)h(x) + f(x)g'(x) h(x) + f(x) g(x) h'(x)$$.

Ex 3.3.7 State and prove a rule to compute $$(fghi)'(x)$$, similar to the rule in the previous problem.

Remark 3.3.2 {Product notation} Suppose $$f_1 , f_2 , \ldots f_n$$ are functions. The product of all these functions can be written $$\prod _{k=1 } ^n f_k.$$ This is similar to the use of $$\sum$$ to denote a sum. For example,

$\prod _{k=1 } ^5 f_k =f_1 f_2 f_3 f_4 f_5$

and

$\prod _ {k=1 } ^n k = 1\cdot 2 \cdot \ldots \cdot n = n!.$

We sometimes use somewhat more complicated conditions; for example

$\prod _{k=1 , k\neq j } ^n f_k$

denotes the product of $$f_1$$ through $$f_n$$ except for $$f_j$$. For example,

$\prod _{k=1 , k\neq 4} ^5 x^k = x\cdot x^2 \cdot x^3 \cdot x^5 = x^{11}.$

Ex 3.3.8 The generalized product rule says that if $$f_1 , f_2 ,\ldots ,f_n$$ are differentiable functions at $$x$$ then

${d\over dx}\prod _{k=1 } ^n f_k(x) = \sum _{j=1 } ^n \left(f'_j (x) \prod _{k=1 , k\neq j} ^n f_k (x)\right).$

Verify that this is the same as your answer to the previous problem when $$n=4$$, and write out what this says when $$n=5$$.