Product Rule for Derivatives

Theorem

Let $f \left({x}\right), j \left({x}\right), k \left({x}\right)$ be real functions defined on the open interval $I$.

Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.

Let $f \left({x}\right) = j \left({x}\right) k \left({x}\right)$.

Then:

$f \, ^{\prime} \left({\xi}\right) = j \left({\xi}\right) \, k \,^{\prime} \left({\xi}\right) + j \,^{\prime} \left({\xi}\right) \, k \left({\xi}\right)$.

It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:

$\forall x \in I: f \,^{\prime} \left({x}\right) = j \left({x}\right) \, k \,^{\prime} \left({x}\right) + j \,^{\prime} \left({x}\right) \, k \left({x}\right)$

Using Leibniz's Notation for Derivatives, this can be written as:

$\dfrac{\mathrm d}{\mathrm dx}\left({y \, z}\right) = y \dfrac{\mathrm dz}{\mathrm dx} + \dfrac{\mathrm dy}{\mathrm dx} z$

where $y$ and $z$ represent functions of $x$.

General Result

Let $f_1 \left({x}\right), f_2 \left({x}\right), \ldots, f_n \left({x}\right)$ be real functions all differentiable as above.

Then:

$\displaystyle D_x \left({\prod_{i=1}^n f_i \left({x}\right)}\right) = \sum_{i=1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)$

Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f \,^{\prime} \left({\xi}\right)\)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac {j \left({\xi + h}\right) k \left({\xi + h}\right) - j \left({\xi}\right) k \left({\xi}\right)} h\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac {j \left({\xi + h}\right) k \left({\xi + h}\right) - j \left({\xi + h}\right) k \left({\xi}\right) + j \left({\xi + h}\right) k \left({\xi}\right) - j \left({\xi}\right) k \left({\xi}\right)} h\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \left({j \left({\xi + h}\right) \frac {k \left({\xi + h}\right) - k \left({\xi}\right)} h + \frac {j \left({\xi + h}\right) - j \left({\xi}\right)} h k \left({\xi}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle j \left({\xi}\right) \, k \,^{\prime} \left({\xi}\right) + j \,^{\prime} \left({\xi}\right) \, k \left({\xi}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Note that $j \left({\xi + h}\right) \to j \left({\xi}\right)$ as $h \to 0$ because, from Differentiable Function is Continuous‎, $j$ is continuous at $\xi$.

Proof of General Result

Proof by induction:

For all $n \in \N_{\ge 1}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle D_x \left({\prod_{i=1}^n f_i \left({x}\right)}\right) = \sum_{i=1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)$

$P(1)$ is true, as this just says:

$D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$

Basis for the Induction

$P(2)$ is the case:

$D_x \left({f_1 \left({x}\right) f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) f_2 \left({x}\right) + f_1 \left({x}\right) D_x \left({f_2 \left({x}\right)}\right)$

which has been proved above.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle D_x \left({\prod_{i=1}^k f_i \left({x}\right)}\right) = \sum_{i=1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)$

Then we need to show:

$\displaystyle D_x \left({\prod_{i=1}^{k+1} f_i \left({x}\right)}\right) = \sum_{i=1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D_x \left({\prod_{i=1}^{k+1} f_i \left({x}\right)}\right)\)	\(=\)	\(\displaystyle D_x \left({\left({\prod_{i=1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i=1}^k f_i \left({x}\right)}\right) + D_x \left({\prod_{i=1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the base case
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i=1}^k f_i \left({x}\right)}\right) + \left({\sum_{i=1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)}\right) f_{k+1} \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{i=1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle D_x \left({\prod_{i=1}^n f_i \left({x}\right)}\right) = \sum_{i=1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \ne i} f_j \left({x}\right)}\right)$ for all $n \in \N$

$\blacksquare$

Mnemonic device

$\left({fg}\right)\,' = f g\,' + f\,' g$

$\left({fgh}\right)\,' = f\,' g h + f g\,' h + f g h\,'$

and in general, making sure to exhaust all possible combinations, making sure that there are as many summands as there are functions being multiplied.

Also see

• Derivative of Product of Real Function and Vector-Valued Function
• Derivative of Cross Product of Vector-Valued Functions
• Derivative of Dot Product of Vector-Valued Functions

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $13.7, \ 13.8$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 10.9 \ \text{(ii)}$