Calculus/Product and Quotient Rules

Product Rule
When we wish to differentiate a more complicated expression such as
 * $$h(x)=(x^2+5x+7)(x^3+2x-4)$$

our only way (up to this point) to differentiate the expression is to expand it and get a polynomial, and then differentiate that polynomial. This method becomes very complicated and is particularly error prone when doing calculations by hand. A beginner might guess that the derivative of a product is the product of the derivatives, similar to the sum and difference rules, but this is not true. To take the derivative of a product, we use the product rule.

It may also be stated as
 * $$(f\cdot g)'=f'\cdot g+f\cdot g'$$

or in the Leibniz notation as
 * $$\dfrac{d}{dx}(u\cdot v)=u\cdot\dfrac{dv}{dx}+v\cdot\dfrac{du}{dx}$$

The derivative of the product of three functions is:
 * $$\dfrac{d}{dx}(u\cdot v \cdot w)=\dfrac{du}{dx}\cdot v\cdot w+u\cdot\dfrac{dv}{dx}\cdot w+u\cdot v\cdot\dfrac{dw}{dx}$$.

Since the product of two or more functions occurs in many mathematical models of physical phenomena, the product rule has broad application in physics, chemistry, and engineering.

Examples

 * Suppose one wants to differentiate $$f(x)=x^2\sin(x)$$ . By using the product rule, one gets the derivative $$f'(x)=2x\sin(x)+x^2\cos(x)$$ (since $$\frac{d}{dx}(x^2)=2x$$ and $$\frac{d}{dx}(\sin(x))=\cos(x)$$).
 * One special case of the product rule is the constant multiple rule, which states: if $$c$$ is a real number and $$f(x)$$ is a differentiable function, then $$c\cdot f(x)$$ is also differentiable, and its derivative is $$(c\cdot f)'(x)=c\cdot f'(x)$$ . This follows from the product rule since the derivative of any constant is 0. This, combined with the sum rule for derivatives, shows that differentiation is linear.

Physics Example I: electromagnetic induction
Faraday's law of electromagnetic induction states that the induced electromotive force is the negative time rate of change of magnetic flux through a conducting loop.
 * $$\mathcal{E}=-\frac{d\Phi_B}{dt}$$

where $$\mathcal{E}$$ is the electromotive force (emf) in volts and &Phi;B is the magnetic flux in webers. For a loop of area, A, in a magnetic field, B, the magnetic flux is given by
 * $$\Phi_B=B\cdot A\cdot\cos(\theta)$$

where θ is the angle between the normal to the current loop and the magnetic field direction.

Taking the negative derivative of the flux with respect to time yields the electromotive force gives
 * $$\begin{align}\mathcal{E}&=-\frac{d}{dt}\big(B\cdot A\cdot\cos(\theta)\big)\\&=-\frac{dB}{dt}\cdot A\cos(\theta)-B\cdot\frac{dA}{dt}\cos(\theta)-B\cdot A\frac{d}{dt}\cos(\theta)\\ \end{align}$$

In many cases of practical interest only one variable (A, B, or θ) is changing, so two of the three above terms are often 0.

Physics Example II: Kinematics
The position of a particle on a number line relative to a fixed point O is $$4t^3\sin(t)\sec^2(t)$$, where $$t$$ represents the time. What is its instantaneous velocity at $$t=7$$ relative to O? Distances are in meters and time in seconds.

Note: To solve this problem, we need some 'tools' from the next section.
 * Answer

We can simplify the function to $$4t^3\tan(t)\sec(t)$$ because ($$\sin(t)\sec(t)=\tan(t)$$)
 * $$v(t)=\frac{d}{dt}\Big[4t^3\tan(t)\sec(t)\Big]=\tan(t)\sec(t)\cdot\frac{d}{dt}[4t^3]+4t^3\sec(t)\cdot\frac{d}{dt}[\tan(t)]+4t^3\tan(t)\cdot\frac{d}{dt}[\sec(t)]$$
 * $$=12t^2\tan(t)\sec(t)+4t^3\sec^3(t)+4t^3\tan^2(t)\sec(t)$$

Substituting $$t=7$$ into our velocity function:
 * $$v(7)=12(7)^2\tan(7)\sec(7)+4(7)^3\sec^3(7)+4(7)^3\tan^2(7)\sec(7)=1496.72\ \frac{m}{s}$$ (to 2 decimal places).

Proof of the Product Rule
Proving this rule is relatively straightforward; first let us state the equation for the derivative:
 * $$\frac{d}{dx}\Big[f(x)\cdot g(x)\Big]=\lim_{h\to0}\frac{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}{h}$$

We will then apply one of the oldest tricks in the book—adding a term that cancels itself out to the middle:
 * $$\frac{d}{dx}\Big[f(x)\cdot g(x)\Big]=\lim_{h\to0}\frac{f(x+h)\cdot g(x+h)-{\color{blue}f(x+h)\cdot g(x)+f(x+h)\cdot g(x)}-f(x)\cdot g(x)}{h}$$

Notice that those terms sum to 0, and so all we have done is add 0 to the equation. Now we can split the equation up into forms that we already know how to solve:
 * $$\frac{d}{dx}\Big[f(x)\cdot g(x)\Big]=\lim_{h\to0}\left[\frac{f(x+h)\cdot g(x+h)-f(x+h)\cdot g(x)}{h}+\frac{f(x+h)\cdot g(x)-f(x)\cdot g(x)}{h}\right]$$

Looking at this, we see that we can factor the common terms out of the numerators to get:
 * $$\begin{align}

\frac{d}{dx}\Big[f(x)\cdot g(x)\Big]&=\lim_{h\to0}\left[f(x+h)\cdot\frac{g(x+h)-g(x)}{h}+g(x)\cdot\frac{f(x+h)-f(x)}{h}\right]\\ &=\lim_{h\to0}\left[f(x+h)\cdot\frac{g(x+h)-g(x)}{h}\right]+\lim_{h\to0}\left[g(x)\cdot\frac{f(x+h)-f(x)}{h}\right]\\ &=\lim_{h\to 0}f(x+h)\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}+\lim_{h\to 0}g(x)\cdot\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{align}$$ Which, when we take the limit, becomes:
 * $$\frac{d}{dx}\Big[f(x)\cdot g(x)\Big]=f(x)\cdot g'(x)+f'(x)\cdot g(x)$$, or the mnemonic "one D-two plus two D-one"

This can be extended to 3 functions:
 * $$\frac{d}{dx}[f\cdot g\cdot h]=f(x)g(x)h'(x)+f(x)g'(x)h(x)+f'(x)g(x)h(x)$$

For any number of functions, the derivative of their product is the sum, for each function, of its derivative times each other function.

Back to our original example of a product, $$h(x)=(x^2+5x+7)(x^3+2x-4)$$, we find the derivative by the product rule is
 * $$h'(x)=(x^2+5x+7)(3x^2+2)+(2x+5)(x^3+2x-4)=5x^4+20x^3+27x^2+12x-6$$

Note, its derivative would not be
 * $${\color{red}(2x+5)(3x^2+2)=6x^3+15x^2+4x+10}$$

which is what you would get if you assumed the derivative of a product is the product of the derivatives.

To apply the product rule we multiply the first function by the derivative of the second and add to that the derivative of first function multiply by the second function. Sometimes it helps to remember the phrase "First times the derivative of the second plus the second times the derivative of the first."

Generalisation
Leibniz gave the following generalisation for the nth derivative of a product;


 * $$(f(x)g(x))^{(n)} = \sum_{i=0}^n \left(\begin{matrix}n\\i\end{matrix}\right)f^{(n-i)}(x)g^{(i)}(x)$$

Where $$\left(\begin{matrix}n \\i\end{matrix}\right)$$ is the binomial coefficient, which may also be written as $$n \text{ choose } i$$ or $${}_nC_i$$.

=Quotient Rule= There is a similar rule for quotients. To prove it, we go to the definition of the derivative:


 * $$\begin{align}

\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]&=\lim_{h\to0}\dfrac{\dfrac{f(x+h)}{g(x+h)}-\dfrac{f(x)}{g(x)}}{h}\\\\ &=\lim_{h\to0}\frac{f(x+h)\cdot g(x)-f(x)\cdot g(x+h)}{h\cdot g(x)\cdot g(x+h)}\\\\ &=\lim_{h\to0}\frac{f(x+h)\cdot g(x)-{\color{blue}f(x)\cdot g(x)+f(x)\cdot g(x)}-f(x)\cdot g(x+h)}{h\cdot g(x)\cdot g(x+h)}\\\\ &=\lim_{h\to0}\dfrac{g(x)\cdot\dfrac{f(x+h)-f(x)}{h}-f(x)\cdot\dfrac{g(x+h)-g(x)}{h}}{g(x)\cdot g(x+h)}\\\\ &=\dfrac{\lim\limits_{h\to0}\left[g(x)\cdot\dfrac{f(x+h)-f(x)}{h}-f(x)\cdot\dfrac{g(x+h)-g(x)}{h}\right]}{\lim\limits_{h\to0}\Big[g(x)\cdot g(x+h)\Big]}\\\\ &=\dfrac{\lim\limits_{h\to0}\left[g(x)\cdot\dfrac{f(x+h)-f(x)}{h}\right]-\lim\limits_{h\to0}\left[f(x)\cdot\dfrac{g(x+h)-g(x)}{h}\right]}{\lim\limits_{h\to0}\Big[g(x)\cdot g(x+h)\Big]}\\\\ &=\dfrac{\lim\limits_{h\to0}g(x)\cdot\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}-\lim\limits_{h\to0}f(x)\cdot\lim\limits_{h\to0}\dfrac{g(x+h)-g(x)}{h}}{\lim\limits_{h\to0}g(x)\cdot\lim\limits_{h\to0}g(x+h)}\\\\ &=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} \end{align}$$

This leads us to the so-called "quotient rule":

Some people remember this rule with the mnemonic "low D-high minus high D-low, square the bottom and away we go!"

Examples
The derivative of $$\frac{4x-2}{x^2+1}$$ is:
 * $$\begin{align}

\frac{d}{dx}\left[\frac{4x-2}{x^2+1}\right]&=\frac{(4)(x^2+1)-(2x)(4x-2)}{(x^2+1)^2}\\ &=\frac{(4x^2+4)-(8x^2-4x)}{(x^2+1)^2}\\ &=\frac{-4x^2+4x+4}{(x^2+1)^2} \end{align}$$

Remember: the derivative of a product/quotient is not the product/quotient of the derivatives. (That is, differentiation does not distribute over multiplication or division.) However one can distribute before taking the derivative. That is $$\frac{d}{dx}\big((a+b)\times(c+d)\big)=\frac{d}{dx}(ac+ad+bc+bd)$$