Calculus (VIII): How To Learn Sum Rule And Difference Rule - An illustration showing the following information using white board graphics: u = f(x); v = f(x); y = f(x) = (u + v); → dy/dx = ??

The sum rule and the difference rule are two fundamental calculus tools that enable us to differentiate combinations of functions. Welcome to the eighth entry in the calculus series. Until now, we have been dealing with functions that are simple algebraic expressions.

As the next step, we are going to learn how to differentiate more complex functions that are additive or subtractive combinations of two or more simpler functions. Without any further ado, let us begin.

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Dealing with a Function that is a Sum of Simpler Functions

Let us say that ‘u’ and ‘v’ are each a function of ‘x’, and ‘y’ is also a function of ‘x’ such that:

u = f(x)

v = f(x)

y = f(x) = u + v

How can we go about differentiating ‘y’ with respect to ‘x’? You might be thinking, “Why not differentiate ‘u’ and ‘v’ separately with respect to ‘x’ and add the results?”

Well, spoiler alert: that is precisely what the sum rule of calculus tells us to do. However, we need to treat the situation formally. That way, we can understand the first principles behind WHY the sum rule works.

For starters, recalling from my previous essays, let us say that we increase ‘x’ by an infinitesimal amount of ‘dx’. As a result, ‘y’ also correspondingly increases by an infinitesimal amount of ‘dy’. This is because ‘y’ is dependent on the value of ‘x’.

But this time, ‘y’ is a more complex function. Its components ‘u’ and ‘v’ are also dependent variables of ‘x’. So, when ‘x’ increases by ‘dx’, not only does ‘y’ increase by ‘dy’, but ‘u’ increases by ‘du’ and ‘v’ increases by ‘dv’. So, the resulting equation is as follows:

y + dy = (u + du) + (v + dv)

The Sum Rule and the Difference Rule Emerge

Now that we have an equation we can work with, let us follow the same procedure we have been following thus far (refer to my previous essays to understand what I mean if you haven’t already):

Calculus (VIII): How To Learn The Sum Rule And The Difference Rule — An illustration showing the following information using white board graphics: (y + dy) = (u + du) + (v + dv); Subtracting (y = u + v) from both sides of the equation: y + dy − y = (u + du) + (v + dv) − (u + v); dy = du + dv
Equation for dy — Math illustrated by the author

Dividing by ‘dx’ throughout, we arrive at the derivative we are looking for:

Calculus (VIII): How To Learn The Sum Rule And The Difference Rule — An illustration showing the following information using white board graphics: dy = du + dv; Dividing both sides by ‘dx’: dy/dx = du/dx + dv/dx
The Derivative — Math illustrated by the author

There we go. Now, we know using first principles as to WHY the sum rule works. Implicitly, this result also conveys the difference rule. If you cannot see how, do not fret. Let us see how using practical examples.

Examples of the Sum Rule and the Difference Rule

Let us first cover the conventional sum rule and then proceed to the difference rule. Consider the following functions:

u = 5x²

v = 3x

y = u + v

To differentiate ‘y’ with respect to ‘x’, we can apply the power rule of calculus that we have already covered to differentiate ‘u’ and ‘v’ first. Following this, we can apply the sum rule to obtain the derivative (dy/dx) as follows:

Calculus (VIII): How To Learn The Sum Rule And The Difference Rule — u = 5x², and v = 3x; Applying the power rule of calculus d(x^n)/dx = nx^(n−1): du/dx = 10x; dv/dx = 3; Applying the sum rule of calculus to y: dy/dx = du/dx + dv/dx = 10x + 3
The sum rule of calculus — Math illustrated by the author

There we go; a practical example of how the sum rule works. No, let us shift our focus to the difference rule. It is a ridiculously simple extension in logic if you consider the following functions:

u = 5x²

v = 3x

z = u − v

To differentiate ‘z’ with respect to ‘x’, we can apply the difference rule as follows:

dz/dx = du/dx − dv/dx

To understand how we arrive at this, all we need to do is modify the original function as follows:

z = u − v = u + (−v)

When we apply the sum rule to this equation, we arrive at the result of the difference rule:

dz/dx = du/dx + d(−v)/dx = du/dx − dv/dx = 10x − 3

Final Thoughts

As I have mentioned in my previous essays, the goal here is to understand why things work the way they do in calculus. So, if you feel that we are dealing with trivial matters here, just know that we are just carving the fundamental building blocks now.

As we proceed further in the series, we will be dealing with incrementally more complex concepts. I hope you enjoy the journey. Up next, I plan to cover the product rule and the quotient rule.

Reference and credit: Silvanus Thompson.

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