Derivative formulas
From Applied Science
Often calculus teachers skip all these proofs below because there isn't enough time to do them all in class.
- [math]\displaystyle{ f(x) = c }[/math]. This is the most trivial derivative: [math]\displaystyle{ \lim_{x \ \to \ h} \frac{f(x + h) - f(x)}{h} = \frac{c - c}{h} = 0 }[/math]. A constant function never changes its value. Therefore its rate of change is zero everywhere.
- [math]\displaystyle{ f(x) = x^n \implies f'(x) = nx^{n \ - \ 1} }[/math]. One confusion that happens here is caused by the tangent line problem. When we differentiate a second degree polynomial, the resulting function is a first degree polynomial and a straight line. However, when the degree is 3 or higher, we still have the tangent line problem, but the derivative won't be a polynomial of degree equal to one!
[math]\displaystyle{ f(x) = x^{-n} \implies f'(x) = -nx^{-n \ - \ 1} }[/math].
[math]\displaystyle{ f(x) = x^{\frac{1}{n}} \implies f'(x) = \frac{1}{n}x^{n \ - \ \frac{1}{n}} }[/math].
- [math]\displaystyle{ (f \pm g)'(x) = f'(x) \pm g'(x) }[/math]. The derivative of the sum is the sum of the derivatives. The proof is identical to the sum of limits, because a derivative is a limit.
- [math]\displaystyle{ (c \cdot f)'(x) = c \cdot f'(x) }[/math]. The same rule for limits.
- [math]\displaystyle{ (f \cdot g)'(x) = f'(x)g(x) + f(x)g'(x) }[/math]. The derivative of the product is not the product of the derivatives! This is contrary to the previous two. With regular limits we have the product of the limits because we have a fixed point. With derivatives, however, we have rates of change, not a fixed point. The geometrical interpretation is similar to that of [math]\displaystyle{ (a + b)^2 }[/math]:
Let's call [math]\displaystyle{ f(x) = u }[/math] and [math]\displaystyle{ g(x) = v }[/math]. Let's assume that their respective values are positive and interpret the product as an area of a rectangle. Going beyond, let's think on differentials and small increments to the variable [math]\displaystyle{ x }[/math]. (Remember that [math]\displaystyle{ |f(a) - f(b)| \gt 0 }[/math]):
[math]\displaystyle{ \Delta u = f(x + \Delta x) - f(x) }[/math]
[math]\displaystyle{ \Delta v = g(x + \Delta x) - g(x) }[/math]
Now, for a small variation in the area of the larger rectangle we have:
[math]\displaystyle{ \Delta (uv) = (u + \Delta u)(v + \Delta v) - uv = u \Delta v + v \Delta u + \Delta u \Delta v }[/math] (the negative term comes from the fact that piece of the area is being counted twice)
Now we want to use the idea of a quotient in the same way Leibniz did, which is also the idea of differentiating a function in respect to its independent variable:
[math]\displaystyle{ \frac{\Delta (uv)}{\Delta x} = u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x} }[/math]
The derivative is a limit, therefore:
[math]\displaystyle{ \lim_{\Delta x \ \to \ 0}\frac{\Delta (uv)}{\Delta x} = \lim_{\Delta x \ \to \ 0}\left( u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x}\right) }[/math]
The third term goes to zero and the other two are a sum of limits, where each term is a derivative under Leibniz's notation:
[math]\displaystyle{ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} }[/math]
Substitute [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] by [math]\displaystyle{ f(x) }[/math] and [math]\displaystyle{ g(x) }[/math] and we have proven the formula.
Note: Can we do [math]\displaystyle{ x^5 = x^3 x^2 }[/math]? Yes, we can use the properties of powers to see one function as a product of two functions. But that won't help in solving derivatives and will only make calculations take unnecessarily longer to perform. In fact, we can see the product of a function by a constant as a special case of the product of two functions.
- [math]\displaystyle{ \left[\frac{f(x)}{g(x)}\right] ' = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} }[/math]. As seen before, the derivative of the quotient is not the quotient of the derivatives. It's very common for people to memorize this rule by memorizing the previous, swapping the plus sign with a minus and dividing everything by the square of the second function. The quickest proof is to use the previous proof, which bypasses the need to write long calculations with limits.
[math]\displaystyle{ \frac{f(x)}{g(x)} = h(x) \iff f(x) = h(x)g(x) }[/math] (In other words, the quotient can be rearranged as a product because there isn't a division by zero)
[math]\displaystyle{ f'(x) = h'(x)g(x) + g'(x)h(x) }[/math]
[math]\displaystyle{ f'(x) - g'(x)h(x) = h'(x)g(x) }[/math]
[math]\displaystyle{ \frac{f'(x) - g'(x)h(x)}{g(x)} = h'(x) }[/math] (to complete the proof we substitute [math]\displaystyle{ h(x) }[/math])
[math]\displaystyle{ \frac{f'(x) - g'(x)\frac{f(x)}{g(x)}}{g(x)} = h'(x) }[/math] (use the least common multiple)
[math]\displaystyle{ \frac{f'(x)g(x) - g'(x)f(x)}{[g(x)]^2} = h'(x) }[/math]
Note: textbooks often omit the [math]\displaystyle{ (x) }[/math] from the proofs because we are doing calculations with functions which are already differentiable and continuous. We don't need to care about the points themselves.
Links for the proofs:
