Derivative of Products & Quotients

Derivative of Product and Quotient

The derivatives are widely used throughout Mathematics and the Sciences, and the Identities simplifying computations are invaluable tools. In this post, our intent is to derive the multiplication and division identities of derivatives using the very first principles.

Derivative

Suppose there be a continuous real function ` y = f(x) ` defined over an open interval `(a, b) \subseteq \mathbb{R}`. Then the derivative of the function `f(x)` at some point `c \in (a, b)` represented by,

`f'(c) or \frac{d}{dx}f(x)` at `x = c`

gives the rate of change of the function at the point `c` with respect to the independent variable. This idea however is still vague and we need a precise mathematical definition.  

Right-hand & Left-hand Rate:

In the neighbour of `c`, let us define an interval `I = [c - h, c + h]` for some ` h > 0`. Then we may define the right-hand average rate of change of the function in the interval as, 

`\frac{f(c+h) - f(c)}{(c+h) - c}`

fig1: The average right-hand change

Similarly we may define the left-hand average rate of change as, 

`\frac{f(c) - f(c-h)}{c - (c-h)}`

fig2: The average left hand change

It should me particularly marked that each time the rate of change is measured with respect to an increasing change in the dependent variable, i.e. we measure the change of the function as we go from left to right on the real number line. So the values shown in the images are not magnitudes but rather algebraic quantities. The average change is equivalent to the slope of the line drawn by joining points `f(c)` and `f(c-h)` (or `f(c+h)`) as required. 

As the size of this interval decrease the average rates of change more accurately describe the rate of change near `c`. And as `h \rightarrow 0`, these average rates of change approximate the instantaneous rate of change near `c`. We then by passing to the limit of `h \rightarrow 0`, define,

` f'_{+}(c) = \lim_{h \rightarrow 0} \frac{f(c+h) - f(c)}{h}`

And,  

` f'_{-}(c) = \lim_{h \rightarrow 0} \frac{f(c) - f(c-h)}{h} `

As the right-hand and left-hand derivates at `x = c` respectively. 

As we had chosen `h > 0`, it is more accurate to write `lim{h \rightarrow 0^{+}}` instead of simply writing `lim{h \rightarrow 0}`, but this is understood.

Actually the above formulas can be written in a more uniform way. The left hand end point of `I` is `lim{h \rightarrow 0^{+}} c - h ` which may be equivalentlly be written as `lim{h \rightarrow 0^{-}} c + h `. This is due to the property that,

 `lim_{h \rightarrow 0^{+}} - h = lim_{h \rightarrow 0^{-}} h`. 

Then the right-hand and left hand derivatives transform as[1],

`f'_{+}(c) = lim_{h \rightarrow 0^{+}} \frac{f(c+h) - f(c)}{h}`

`f'_{-}(c) = lim_{h \rightarrow 0^{-}} \frac{f(c+h) - f(c)}{h}`

When both the left-hand and right-hand derivatives exist finitely and are equal, then we say the derivative at `x = c` exists and is equal to, 

  $$ \boxed{f(c) = f'_{+}(c) = f'_{-}(c)} $$ 

The following interactive graph shows us the variation of right-hand and left-hand as `h` becomes smaller and smaller around an arbitrarily chosen point `c`. Use your mouse or fingers to drag the points. You can drag any of the point, and the graph adjusts itself!

Fig: Interactive graph showing the variation RHR and LHR as `h \rightarrow 0` . See that near `c` , LHR and RHR become almost equal suggesting that the derivative does exist and its slope equal to either of  LHR or RHR. Use your fingers to drag the points.

This point `c` can be any point `x`, in the neighborhood of which the function is continuous. Thus, the derivative of the function `f(x)` in the general form can be written as,

$$ \boxed{f'(x) = lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}} $$

The derivatives exist when this limit exists. This is the first principle definition of the derivative.

Product of two functions:

Let, `f(x)` and `g(x)` be two real functions continuous in some open interval `(a, b) \subseteq (D_{f} \cap D_{g})` . Let's define the product of the two functions as `ψ(x) = f(x) \cdot g(x) ` in `(a, b)`. Then its derivative,

`ψ'(x) = \lim _{h \to 0} \frac{ψ(x+h) - ψ(x)}{h} `

` = lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h} `

In the first term, we can generate `f'(x)` if we have a `- f(x)` term. For this we subtract `f(x)g(x+h)`. To compensate, we also add `f(x)g(x+h)`. Then the above becomes, 

`\lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h} `

` = \lim_{h \to 0} \frac{g(x)[f(x+h) - f(x)] + f(x)[g(x+h) - g(x)]}{h} `

` = g(x) \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} + f(x) \cdot \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} `

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

Therefore, 

$$ \boxed{ \frac{d}{dx} f(x)g(x) = g(x)f'(x) + f(x)g'(x)} $$

Division of two functions:

Similar to above, now let `ψ(x) = \frac{f(x)}{g(x)}`, with ` g(x) \neq 0` in `(a, b)`. Then, 

`ψ'(x) = \lim_{h \to 0} \frac{ψ(x +h) - ψ(x)}{h} `

`= lim_{h \to 0} \frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}}{h}`

Cross multiplying, the above becomes,

`\lim_{h \to 0} \frac{f(x+h)g(x) - f(x)g(x+h)}{h g(x+h)g(x)`

To generate `f'(x)` we will need to subtract `f(x)g(x)` and consequently will also need to add the same. The above becomes,

`lim_{h \to 0} \frac{f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h)}{hg(x+h)g(x)}`

`lim_{h \to 0} \frac{g(x)[f(x+h) - f(x)] - f(x)[g(x+h) - g(x)]}{hg(x+h)g(x)} `

Using the properties of limits this can be transformed as,

`lim_{h  \to 0} frac{1}{g(x)g(x+h)}[g(x) \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} - f(x) \cdot \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}]`

`= \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^{2}}`

Therefore, 

$$ \boxed{ \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^{2}} } $$

Thus, we have derived the identities for the product and quotient of two functions.

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Notes:

[1] The right hand derivative is straight forward. The left hand derivative was precisely, 

` \lim_{h \to 0^{+}} \frac{f(c) - f(c - h)}{h} `

= ` \lim_{h \to 0^{+}} \frac{f(c + (- h)) - f(c)}{-h} `

Then we use the property that we earlier mentioned that `lim_{h \to 0^{+}} -h = lim_{h \to 0^{-}} h`. So that, 

`f'_{-}(c) =  \lim_{h \to 0^{+}} \frac{f(c + (- h)) - f(c)}{-h} `

` =  \lim_{h \to 0^{-}} \frac{f(c + h) - f(c)}{h} `

which is as required. However instead of manipulation, I like to interpret this more intuitively. That when `h` is slightly less than `0` it is negative, and adding it will slightly decrease the value of term.

Writing the derivative in this form is also a matter of personal preference. Some readers may feel close to the previous form of using `h \to 0^{+}`, but we should note that both definitions are equivalent.