1. Constant rule Let’s continue our introduction to derivatives with some basic, yet incredibly handy, properties for differentiation. The first is called the constant rule. The rule basically says that when a function is a number times another function, we can essentially ignore that number for derivative purposes. Here it is more explicitly.

Theorem 1.1 (Constant rule for derivatives). Let f(x) = c⋅g(x), where c is some real number. Then f ′(x) = c ⋅ g′(x).

That’s the whole theorem. This will become second nature to you very quickly. The proof is short and sweet, so let’s have a look at it.

Proof. We begin with the definition of the derivative.

f(x + ∆x) − f(x + ∆x) c ⋅ g(x + ∆x) − c ⋅ g(x) f ′(x) = lim = lim ∆x→0 ∆x ∆x→0 ∆x g(x + ∆x) − g(x) = lim c ⋅ = c ⋅ g′(x). ∆x→0 ∆x 

That’s the end of the proof - our shortest one yet. Let’s see a couple examples.

Example 1. Find the derivatives of f(x) = 4x8 and g(x) = −4~3x−1.

By the constant rule and the power rule, we have f ′(x) = 32x7 and g′(x) = ~ −2 = 4 4 3x 3x2 .

2. Addition rule Suppose f(x) = x2 + x3. What would we like f ′(x) to be? It would be great if f ′(x) = 2x + 3x2. In fact, it is! It turns out that differentiation additively distributes. We make this explicit in the following theorem :

Theorem 2.1 (Addition rule for derivatives). Suppose f(x) is a function that is of the form f(x) = g(x)+h(x), where g(x) and h(x) are arbitrary functions. Then f ′(x) = g′(x) + h′(x). 1 2

The proof of the theorem is quite simple, so it is worth it to see how it is done.

Proof. We begin with the definition of the derivative.

f(x + ∆x) − f(x) g(x + ∆x) + h(x + ∆x) − (g(x) + h(x) f ′(x) = lim = lim ∆x→0 ∆x ∆x→0 ∆x g(x + ∆x) − g(x) + h(x + ∆x) − h(x) = lim ∆x→0 ∆x g(x + ∆x) − g(x) h(x + ∆x) − h(x) = lim + ∆x→0 ∆x ∆x g(x + ∆x) − g(x) h(x + ∆x) − h(x) = lim + lim = g′(x) + h′(x). ∆x→0 ∆x ∆x→0 ∆x 

This is such a nice theorem that you will quickly be using it nearly everyday in this class and won’t even realize it. It will become just that natural to you. Let’s see some examples.

Example 2. Find the derivative of f(t) = t3 + t5.

From the addition rule and power rule, we have f ′(t) = 3t2 + 5t2.

Example 3. Find the derivative of g(x) = 4t−2 − 7.

′( ) = − −3 − = − 8 From the addition, constant, and power rules, we have g x 8x 0 x3 .

3. Product rule Suppose f(x) = x2 ⋅ sin x. What would nice for the derivative to be? It would be great if f ′(x) = 2x ⋅ ( whatever the derivative of sin x is. However, it is not. It is a bit more complicated than that. Luckily, there still is a nice theorem for when we consider products of functions, it’s just not what we dream it would be.

Theorem 3.1 (Product rule for derivatives). Let f(x) be a product of functions. That is, let f(x) = g(x) ⋅ h(x). Then f ′(x) = g(x) ⋅ h′(x) + g′(x) ⋅ h(x). 3

We will not prove the product rule as it is a homework assignment. It goes in a very similar fashion to those of the constant rule and the addition rule. We also don’t have any examples right now as all of our functions are composed only of power functions, which do not make for interesting examples for the product rule.

4. Chain rule The chain rule. Many students have a love-hate relationship with the chain rule. It is a bit slippery at first, but with some practice, it becomes a wonderful and effective tool. The chain rule works with compositions of functions. For example, if f(x) = sin(x3), we may wonder what f ′(x) is. The chain rule will tell us.

Theorem 4.1 (Chain rule for derivatives). Let f(x) be a composition of functions. That is, let f(x) = g(h(x)). Then f ′(x) = g′(h(x)) ⋅ h′(x). In a different notation, ∂f ∂g ∂h = ⋅ . ∂x ∂h ∂x

An easy way to remember this is, “derivative of the outside times the derivative of the inside.” That is, we find the derivative of the outside function. Next, we plug the original inside function into this derivative. Finally, we multiply everything by the derivative of the inside.

( ) = 1 Example 4. Find the derivative of f x x2−4 .

1 2 − The “outside” function is x . The “inside” function is x 4. Thus, we have ′( ) = − 1 ′( ) = ′( ) = − 1 ⋅ = − 2x g x x2 and h x 2x. So, the derivative is f x (x2−4)2 2x (x2−4) .

Example 5. Find the derivative of f(x) = (3x4 − x3)97.

The “ouside” fuction is g(x) = x97. The “inside” function is h(x) = (3x4 − x3). So g′(x) = 97x96 and h′(x) = 12x3 − 3x2. So by the chain rule, we have f ′(x) = 97(3x4 − x3)96 ⋅ (12x3 − 3x2).