Product Rule for Derivatives

Suppose we have a function we don't know how to take the derivative of, but the function is the product of two functions we do know how to take the derivative of, like this:

$h(x) = (x^{2} + 3x + 6)(x^{8} + x)$

It would be nice to have a formula for the derivative of $f(x)$ in terms of the derivatives of the two functions on the right. Let's call these functions $f(x) = x^{2} + 3x + 6$ and $g(x) = x^{8} + x$.

To do this, imagine a rectangle with sides $f(x)$ and $g(x)$, and suppose $x$ represents time. Then $h(x)$ is the area of the rectangle at time $x$:

Now suppose a small amount of time, $\Delta x$, passes. The values of the two functions will also change by small amounts, $\Delta f$ and $\Delta g$, and the rectangle's dimensions will change:

We want the derivative of $h(x)$, so let's find $\Delta h(x)$, the difference between the area of the red rectangle and the area of the blue rectangle.

We're going to find the difference by adding the areas of the three rectangles formed by the dotted red lines. (If you have a different strategy, try using yours and see if you get the same thing!)

$\Delta h(x) = f(x)\Delta g(x) + g(x)\Delta f(x) + \Delta f(x)\Delta g(x)$

Divide both sides of the equation above by $\Delta x$; this takes us from an equation about differences, to an equation about differences relative to a small change in $x$.

$\dfrac{\Delta h(x)}{\Delta x} = \dfrac{f(x)\Delta g(x)}{\Delta x} + \dfrac{g(x)\Delta f(x)}{\Delta x} + \dfrac{\Delta f(x)\Delta g(x)}{\Delta x}$

Now we take the limit of both sides as $\Delta x$ approaches 0. (Yes, this is allowed; since the two sides are always equal, they must have the same limit!)

If you took the limit of both sides explicitly and used limit notation, you should have:

$h'(x) = f(x)g'(x) + g(x)f'(x) + \displaystyle\lim_{\Delta x\rightarrow 0}\dfrac{\Delta f(x)\Delta g(x)}{\Delta x}$

If you didn't use limit notation, but you're working under the assumption that $\Delta x$ is approaching 0, you should have:

$h'(x) = f(x)g'(x) + g(x)f'(x) + \dfrac{\Delta f(x)\Delta g(x)}{\Delta x}$

Now for the final step:

If you followed the hints, you've seen most of the solution:

As $\Delta x$ approaches 0, we can rewrite the term as $f'(x)\Delta g(x)$. Then $\Delta g(x)$ approaches 0, so the whole term approaches 0, leaving us with:

$h'(x) = f(x)g'(x) + g(x)f'(x)$

And that's the product rule!

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Another way to think about why the last term disappears as $\Delta x$ approaches 0 is to think about the areas of the rectangles. The area of the small rectangle in the upper right becomes insignificant compared to the other two rectangles.

Finally, let's solve the problem we started with.

We'll finish by writing out the product rule one more time:

If $h(x) = f(x)g(x)$, then $h'(x) = f(x)g'(x) + g(x)f'(x)$.

I often like limiting cases when I derive equations. What we want to know is a way to take the derivative of the product of two functions

$$\frac{d}{dx} (f(x) g(x))$$

in terms of $f'(x)$ and $g'(x)$. Well, if we got lucky and $f(x) = f_0$ was a constant this would be easier.

$$\frac{d}{dx} (f_0 g(x)) = f_0 g'(x)$$

Similarly, if $g(x)=g_0$ was a constant function, we would have gotten $g_0 f'(x)$ for the derivative. These cases will let us know when we get the right answer.

What if we estimate $f(x) \approx f_0 + \Delta x\;f'$ and $g(x) = g_0 + \Delta x\;g'$. Then,

$$\frac{d}{dx} (f(x) g(x))= f_0 g'(x)\\ = \frac{d}{dx} \left[(f_0+ \Delta x \;f') (g_0+ \Delta x \;g') \right]\\ \approx \frac{d}{dx} \left( f_0 g_0 + \Delta x\;(f_0 g' + f' g_0) \right) \\ \approx f(x) g'(x) + f'(x) g(x)$$

The derivative of product of functions $f(x)$ and $g(x)$ is equal to the product of $f(x)$ and the derivative of $g(x)$ plus the product of the derivative of $f(x)$ and $g(x)$. $$\begin{equation} (f(x) \cdot g(x))'=f(x)\cdot g(x)'+f(x)'\cdot g(x) \end{equation}$$

This is an intuitive explanation of the product rule using a physical example...

Imagine you're tiling a floor with little square tiles that have side length $s$. You have a tiled patch that is 5 tiles long and 3 tiles wide. The length of the patch $(L)$ and width of the patch $(W)$ can be expressed as functions of $s$.

The area you have covered $(A)$ can also be expressed as a function of s, in terms of $L$ and $W$

Now let's consider what happens if we change the size of the tiles by a small amount to $s+\Delta s$

If $\Delta s$ is a very small change in the tile size, then $(\Delta s)^2$ will be even smaller, so that term can be ignored.

As $\Delta s$ becomes infinitesimally small, this will give the derivative of the area:

$A'(s) = \lim_{\Delta s \rightarrow 0} \frac{\Delta A}{\Delta s} = 30 s$

Now let's work out the change in area using the product rule.

And there we have a very simple demonstration of the product rule. This will be more clear once we can draw a basic sketch of the area as the length and width change, to illustrate this example. I highly recommend you try it yourself!

(This explanation uses little-o notation.)

Suppose functions $f(x)$, $g(x)$ are differentiable at $a$. What can we say about $(f(x)g(x))' |_{x=a}$, the derivative of $f(x)g(x)$ evaluated at $x=a$?

Perhaps we can get an expression in terms of $f(a),f'(a),g(a),g'(a)$.

Correct!

You may have come up with your own example, but for instance, consider $f(x) = x$, $g(x) = 1$. We know $(f(x)g(x))' = (x)' = 1$ for all $x$, while $f'(x)g'(x) = 1\cdot0 = 0$ for all $x$. In particular, for every real $a$, we have $(f(x)g(x))' |_{x=a} \ne f'(a)g'(a)$.

Alternatively, if we use "dimensional analysis", then if $f(x),g(x)$ represent distances traveled at time $x$, the units of $(f(x)g(x))'$ are something like $m^2/s$, while the units of $f'(x)g'(x)$ are something like $m^2/s^2$. So it doesn't really make sense for $(f(x)g(x))' |_{x=a}$ to equal $f'(a)g'(a)$. (At least not in general.) If you're interested (otherwise feel free to skip ahead), the dimensional analysis can be formalized like this: Suppose $(f(x)g(x))' = f'(x)g'(x)$ were always true, for any functions $f,g$. Let's "pretend" $f,g$ have units of $m$, and $x$ has units of $s$. Then we can scale $x$ (intuitively, "scale time") as follows: for a positive constant $c$, let $F(x) = f(cx)$ and $G(x) = g(cx)$. (Physically, we're scaling the units of time by a factor of $c$.) Then for any constant $a$, we have $F'(x) |{x=a} = c f'(x) |{x=ca}$, $G'(x) |{x=a} = c g'(x) |{x=ca}$, and $(F(x)G(x))' |{x=a} = c (f(x)g(x))' |{x=ca}$. (Physically, since we scaled time by a factor of $c$, the functions $F,G,F\cdot G$ grow, in some sense, $c$ times as fast as $f,g,f\cdot g$ do.) So if $(f(x)g(x))' = f'(x)g'(x)$ at $x=ca$, then $(F(x)G(x))' = F'(x)G'(x) / c \ne F'(x)G'(x)$ at $x=a$, at least if $F'(a),G'(a)\ne 0$. (Why?)

Well, we have $$f(x)g(x)\approx [f(a) + f'(a)(x-a)] [g(a) + g'(a)(x-a)]\\ = f(a)g(a) + [f(a)g'(a) + f'(a)g(a)](x-a) + (x-a)^2$$ $$f(x)g(x)\in [F(x)+o(x-a)] [G(x)+o(x-a)]$$ $$\subseteq F(x)G(x) + F(x) o(x-a) + G(x) o(x-a) + o((x-a)^2).$$ We already analyzed $F(x)G(x)$ above. We also know $o((x-a)^2)\subseteq o(x-a)$ is negligible. Now let's think about $F(x) o(x-a)$. Since $F(x)$ is bounded by a constant near $x=a$, we simply have $F(x) o(x-a)\subseteq o(x-a)$. Similarly, $G(x) o(x-a)\subseteq o(x-a)$. Thus, $$f(x)g(x)\in F(x)G(x) + o(x-a)$$ $$\subseteq [f(a)g(a) + [f(a)g'(a) + f'(a)g(a)](x-a) + o(x-a)] + o(x-a)$$ $$\subseteq f(a)g(a) + [f(a)g'(a) + f'(a)g(a)](x-a) + o(x-a)$$ since the sum of finitely many $o(x-a)$ terms is still $o(x-a)$.

This resembles a linear approximation for $f(x)g(x)$, but we have an extra $(x-a)^2$ term at the end. It's negligible, though, since $(x-a)^2\in o(x-a)$. So we expect the derivative to be $f(a)g'(a) + f'(a)g(a)$. This turns out to be correct! More formally, letting $F(x) = f(a) + f'(a)(x-a)$ and $G(x) = g(a)+g'(a)(x-a)$, we have the slope of this linear approximation is indeed $f(a)g'(a) + f'(a)g(a)$, as desired. (Is this consistent with dimensional analysis?

For some graphical intuition, see Wikipedia's image.

Although the Product Rule is most commonly seen for the expansion of a two-factor product, if there are more than two factors, the Product Rule can be extended for these more complicated expressions.

$(f(x)g(x)h(x))' = ([f(x)g(x)]\cdot[h(x)])'$

The Product Rule can be used where the factors are $f(x)g(x)$ and $h(x)$:

$$\begin{align*}&(f(x)g(x))'h(x) + (f(x)g(x))h'(x) \\&= h(x)(f'(x)g(x) + f(x)g'(x)) + f(x)g(x)h'(x) \\&= f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x) \end{align*}$$

(This will possibly need some knowledge of combinatorics, induction, and binomial coefficients. This is very optional, and for the very interested reader. :) )

In this post we will generalize the Product-Rule, showing something called the General Leibniz Rule.

Define $f^{(n)}(x)$ to be the result when you differentiate $f(x)$ $n$ times. For example, if $f(x) = x^5, f^{(3)}(x) = 60x^2$. From now on, I will drop the $(x)$ for simplicity, so $f(x) \equiv f$. Define $fg(x) = f(x)g(x)$. The product rule says that $(fg)^{(1)} = f^{(1)}g + fg^{(1)}$.

Computing $(f^{(1)}g + fg^{(1)})' \\= (f^{(1)}g)' + (fg^{(1)})' \\= (f^{(2)}g + f^{(1)}g^{(1)}) + (f^{(1)}g^{(1)} + fg^{(2)})$

$= f^{(2)}g + 2f^{(1)}g^{(1)} + fg^{(2)}.$

Look at the coefficients in $(fg)^{(1)}, (fg)^{(2)}, (fg)^{(3)}$. They are:

1 1

1 2 1

1 3 3 1

Hmmm looks familiar

You can learn about Pascal's Triangle here.

Every entry in Pascal's Triangle is generated by adding the above two entries together.

In fact, the General Leibniz rule says that when you compute $(fg)^{(n)}$, the coefficients are precisely the entries of the $n$-th row of Pascal's triangle! More formally, because the entries in Pascal's triangle are binomial coefficients, the $k$-th entry in the $n$-th row is $\binom{n}{k}$. The formula is: $$(fg)^{(n)} = \sum_{k = 0}^n \binom{n}{k} f^{(k)}g^{(n-k)} .$$

We're now going to finish this up using induction. Though this may sound scary, all we're going to do is take the formula for $(fg)^{(n)}$, differentiate it, and show that we indeed get the formula for $(fg)^{(n+1)}$. Let's start!

Using this formula, we can finish our proof. $$((fg)^{(n)})' = (\sum_{k = 0}^n \binom{n}{k} f^{(k)}g^{(n-k)}) = \sum_{k = 0}^n \binom{n}{k} (f^{(k)}g^{(n-k)})'$$ $$= \sum_{k = 0}^n \binom{n}{k} (f^{(k+1)}g^{(n-k)} + f^{(k)}g^{(n-k+1)}) \\= \sum_{k = 0}^{n+1} \left(\binom{n}{k-1} + \binom{n}{k} \right) f^{(k)}g^{(n+1-k)} .$$

Note that in the above some of the binomials are undefined: for example $\binom{n}{-1}$ appears. Don't worry: as a general rule, if undefined binomial coefficients appear, treat them as $0$. To finish, if we could just show that $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$, we would be done!

Therefore, $$\sum_{k = 0}^{n+1} \left(\binom{n}{k-1} + \binom{n}{k} \right) f^{(k)}g^{(n+1-k)} = \sum_{k = 0}^{n+1} \binom{n+1}{k} f^{(k)}g^{(n+1-k)},$$

and we are done!

Thanks for staying with me to this point! Hopefully all the hyperlinks to Wikipedia can be links to Expii pages in the future, when we release Combinatorics in some form. Some of it will definitely be covered in our Algebra map that will be released soon!

The derivative of a product of two functions is equal to the sum of the first function times the derivative of the second function plus the derivative of the first function times the second function.

$(f(x)g(x))'$ = $f(x)g(x)'$ + $f(x)'g(x)$

PROOF OF PRODUCT RULE:

$\frac{d}{dx} [f(x)g(x)] = \lim \frac{f(x+a)g(x+a) - f(x)g(x)}{a} \\= \lim \frac{f(x+a)g(x+a)-f(x)g(x+a) + f(x)g(x+a) - f(x)g(x)}{a} \\= \lim \frac{g(x+a)(f(x+a) - f(x))}{a} + \lim \frac{f(x)(g(x+a) - g(x))}{a} \\= f'(x)g(x) + f(x)g'(x).$

One might think that the derivative of a product is the product of the derivatives in question. However, this is wrong. An example of this is:

Let $f(x)=x$ and $f(g)=x^2$

$f'(x)=1$ and $f'(g)=2x$

but $(fg)(x)=x^3$ so $(fg)'(x)=3x^2$

Thus $(fg)'$ does not equal $f'g'$ because $f'g'=2x$ not $3x^2$

So how does one actually solve this properly? You use the product rule. The product rule is viewed as such:

So lets reason the problem properly now that we know what to do.

$h(x)=f(x) * g(x)$ where $f(x)=x$ and $g(x) = x^2$

so lets find $h'(x)$...

$h'(x)= 2x(x) + 1(x^2)$ $h'(x)= 2x^2 + x^2$ $h'(x)=3x^2$

and there we have the previous, correct, answer. This is how one does the product rule but remember to always make sure you are getting the proper derivatives from both f(x) and g(x) within the formula.

We know how to find the derivative of functions $f(x)$ and $g(x)$. The derivatives are $f'(x)$ and $g'(x)$ respectively.

But what about the derivative of function $h(x)$ which is composed of functions $f(x)$ and $g(x)$?

To find the derivative of a function $h(x)$ that composes of two functions $f(x)$ and $g(x)$, one must use one of three different rules. The rules are the Product Rule, the Quotient Rule and the Chain Rule.

The Product Rule helps when $h(x)$ is the $product$ of two different functions $f(x)$ and $g(x)$. The rule states that

If $h(x) = f(x)*g(x)$ then $h'(x) = f'(x)*g(x) + f(x)*g'(x)$.

The way I remember this rule is by remembering this rhyme:
"$first \; d \;second\; second \;d \;first$". The $d$ refers to the notation that says $df(x) = f'(x)$.

The Quotient Rule helps when $h(x)$ is the $quotient$ of two different functions $f(x)$ and $g(x)$. The rule states that

If $h(x) = \frac{f(x)}{g(x)}$ then $h'(x) = \frac{g(x)*f'(x) - f(x)*'g(x)}{g^2(x)}$

The way I remember this rule is by remembering this rhyme:
"$low\; d\; high\; minus\; high \; d \; low \; all \; over \; low \; squared$"

The Chain Rule helps when $h(x)$ is a function of a function in a function or a $chain$ function such as $f(g(x))$. The rule states that

If $h(x) = f(g(x))$ then $h'(x) = g'(x)*f'(g(x))$

The way I remember this rule is by remembering that I just take the derivative of the outer function and multiply it by the derivative of the inner.

Here are some extra practice problems, have fun solving them!

(Videos) Using Chain, Product, and Quotient Rules to Find Derivatives

by PatrickJMT

Here's a video by patrickJMT showing you how to use the chain, product, and quotient rules to find the derivatives of functions.

Recall the chain, product, and quotient rules: $$\begin{align} [f(g(x))]'&=f'(g(x))\cdot g'(x)\\ [f(x)g(x)]'&=f'(x)g(x)+g'(x)f(x)\\ \bigg[\frac{f(x)}{g(x)}\bigg]'&=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2} \end{align}$$

The first problem he finds the derivative of is: $$f(x)=(2x^{\frac13}-6x^{\frac53})^3\cdot(4x^2-5x)^{\frac13}$$

I like to calculate each pieces derivative separately and then plug them into the formula after. It helps me keep things organized.

So I'll start by finding the derivative of the first part. $$f_1(x)=(2x^{\frac13}-6x^{\frac53})^3$$ Here's where we use the chain rule. Take the derivative of the outside and then multiply it by the derivative of the inside. $$\begin{align} f_1'(x)&=\color{red}{3}(2x^{\frac13}-6x^{\frac53})^{\color{red}{3-1}}\color{red}{\cdot(2x^{\frac13}-6x^{\frac53})'}\\ &=3(2x^{\frac13}-6x^{\frac53})^2\cdot[2(\color{red}{\frac13}x^{\color{red}{\frac13-1}})-6(\color{red}{\frac53}x^{\color{red}{\frac53-1}})]\\ &=\boxed{3(2x^{\frac13}-6x^{\frac53})^2\cdot\bigg(\frac23x^{-\frac23}10x^{\frac23}\bigg)} \end{align}$$

Now let's find the derivative of the second part. $$f_2(x)=(4x^2-5x)^{\frac13}$$ Again, we need to use the chain rule. $$\begin{align} f_2'(x)&=\color{red}{\frac13}(4x^2-5x)^{\color{red}{\frac13-1}}\color{red}{\cdot(4x^2-5x)'}\\ &=\frac13(4x^2-5x)^{-\frac23}\cdot[4(\color{red}{2}x^{\color{red}{2-1}})-5(\color{red}{1}x^{\color{red}{1-1}})]\\ &=\boxed{\frac13(4x^2-5x)^{-\frac23}\cdot(8x-5)} \end{align}$$

Now that we have all the pieces we can plug them into our product rule for derivatives formula. $$\begin{align} f'(x)&=f_1'(x)\cdot f_2(x)+f_2'(x)\cdot f_1(x)\\ &=3(2x^{\frac13}-6x^{\frac53})^2\cdot\bigg(\frac23x^{-\frac23}10x^{\frac23}\bigg)\cdot(4x^2-5x)^{\frac13}+\\ &\ \ \ \ \ +\frac13(4x^2-5x)^{-\frac23}\cdot(8x-5)\cdot(2x^{\frac13}-6x^{\frac53})^3\\ &=3(2x^{\frac13}-6x^{\frac53})^2\cdot\bigg(\frac23x^{-\frac23}10x^{\frac23}\bigg)\cdot(4x^2-5x)^{\frac13}+\\ &\ \ \ \ \ +\frac{(8x-5)(2x^{\frac13}-6x^{\frac53})^3}{3(4x^2-5x)^{\frac23}} \end{align}$$