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.

Derivatives of products (Product Rule)

${\frac {d}{dx}}{\Big [}f(x)\cdot g(x){\Big ]}=f'(x)\cdot g(x)+f(x)\cdot g'(x)$

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: rocket acceleration

The acceleration of model rockets can be computed with the product rule.

Consider the vertical acceleration of a model rocket relative to its initial position at a fixed point on the ground. Newton's second law says that the force is equal to the time rate change of momentum. If ${\vec {F}}$ is the net force (sum of forces), ${\vec {p}}$ is the momentum, and $t$ is the time,

{\vec {F}}={\frac {d{\vec {p}}}{dt}}

Since the momentum is equal to the product of mass and velocity, this yields

{\vec {F}}={\frac {d}{dt}}(m{\vec {v}})

where $m$ is the mass and $v$ is the velocity. Application of the product rule gives

{\vec {F}}={\vec {v}}{\frac {dm}{dt}}+m{\frac {d{\vec {v}}}{dt}}

Since the acceleration, ${\vec a}$ , is defined as the time rate change of velocity, ${\vec {a}}={\frac {d{\vec {v}}}{dt}}$ ,

{\vec {F}}={\vec {v}}{\frac {dm}{dt}}+m{\vec {a}}

Solving for the acceleration,

{\vec {a}}={\dfrac {{\vec {F}}-{\vec {v}}{\dfrac {dm}{dt}}}{m}}

Since the rocket is losing mass, ${\frac {dm}{dt}}$ is negative, and the changing mass term results in increased acceleration.^[1]^[2]

Note: Here ${\vec {F}}$ is considered to be the net force.

Physics Example II: 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 Φ_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{aligned}{\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{aligned}}

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 III: 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.

Answer

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

We can simplify the function to $4t^{3}\tan(t)\sec(t)$ ( $\sin(t)\sec(t)=\tan(t)$ )

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\to 0}{\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\to 0}{\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\to 0}\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{aligned}{\frac {d}{dx}}{\Big [}f(x)\cdot g(x){\Big ]}&=\lim _{h\to 0}\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\to 0}\left[f(x+h)\cdot {\frac {g(x+h)-g(x)}{h}}\right]+\lim _{h\to 0}\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{aligned}}

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."

Quotient Rule

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

{\begin{aligned}{\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]&=\lim _{h\to 0}{\dfrac {{\dfrac {f(x+h)}{g(x+h)}}-{\dfrac {f(x)}{g(x)}}}{h}}\\\\&=\lim _{h\to 0}{\frac {f(x+h)\cdot g(x)-f(x)\cdot g(x+h)}{h\cdot g(x)\cdot g(x+h)}}\\\\&=\lim _{h\to 0}{\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\to 0}{\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\to 0}\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\to 0}{\Big [}g(x)\cdot g(x+h){\Big ]}}}\\\\&={\dfrac {\lim \limits _{h\to 0}\left[g(x)\cdot {\dfrac {f(x+h)-f(x)}{h}}\right]-\lim \limits _{h\to 0}\left[f(x)\cdot {\dfrac {g(x+h)-g(x)}{h}}\right]}{\lim \limits _{h\to 0}{\Big [}g(x)\cdot g(x+h){\Big ]}}}\\\\&={\dfrac {\lim \limits _{h\to 0}g(x)\cdot \lim \limits _{h\to 0}{\dfrac {f(x+h)-f(x)}{h}}-\lim \limits _{h\to 0}f(x)\cdot \lim \limits _{h\to 0}{\dfrac {g(x+h)-g(x)}{h}}}{\lim \limits _{h\to 0}g(x)\cdot \lim \limits _{h\to 0}g(x+h)}}\\\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}\end{aligned}}

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

Derivatives of quotients (Quotient Rule)

${\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]={\frac {g(x)f'(x)-f(x)g'(x)}{g(x)^{2}}}$

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{aligned}{\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{aligned}}

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 ${\frac {d}{dx}}{\big (}(a+b)\times (c+d){\big )}={\frac {d}{dx}}(ac+ad+bc+bd)$

References

↑ Chandler, David (October 2000). "Newton's Second Law for Systems with Variable Mass". The Physics Teacher 38 (7): 396.
↑ Courtney, Michael; Courtney, Amy. "Measuring thrust and predicting trajectory in model rocketry". arΧiv:0903.1555.

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