MATH 145 – Calculus I
October 6, 2025
Recall the definition of the derivative function.
Let \(f\) be a function and \(x\) a value in the function’s domain. We define the derivative of \(f\), a new function called \(f'\), by the formula \[ f'(x) = \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}, \] provided this limit exists.
We are limited in what can be accomplished from this definition alone, we would like to develop some rules to make things smoother.
For any real number \(c\), if \(f(x) = c\), then \(f'(x) = 0\).
Proof: If \(f(x) = c\), then: \[ \begin{aligned} f'(x) &= \lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim\limits_{h \to 0} \frac{c - c}{h} \\ &= \lim\limits_{h \to 0} \frac{0}{h} \\ &= 0. \end{aligned} \]
Examples:
Find \(f'(x)\) if \(f(x) = 42\).
Find \(g'(x)\) if \(g(x) = e^{110.09} + \log(16.51) - \pi^i\).
For any real number \(k\), if \(f(x)\) is a differentiable function with derivative \(f'(x)\), then \(\dfrac{d}{dx} \left[ k f(x) \right] = k f'(x)\).
Proof: \[ \begin{aligned} \frac{d}{dx} \left[ k f(x) \right] &= \lim\limits_{h \to 0} \frac{k f(x + h) - k f(x)}{h} \\ &= \lim\limits_{h \to 0} \frac{k \left(f(x + h) - f(x)\right)}{h} \\ &= \lim\limits_{h \to 0} k \left(\frac{f(x + h) - f(x)}{h}\right) \\ &= k \left(\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h}\right) \\ &= k f'(x). \end{aligned} \]
Examples:
If \(f(x)\) and \(g(x)\) are differentiable functions with derivatives \(f'(x)\) and \(g'(x)\) respectively, then \(\dfrac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)\).
Proof: \[ \begin{aligned} \frac{d}{dx} \left[ f(x) + g(x) \right] &= \lim\limits_{h \to 0} \frac{\left(f(x+h) + g(x+h) \right) - \left(f(x) + g(x) \right)}{h} \\ &= \lim\limits_{h \to 0} \frac{\left(f(x+h) - f(x) \right) + \left(g(x+h) - g(x) \right)}{h} \\ &= \lim\limits_{h \to 0} \left[ \left( \frac{f(x+h) - f(x) }{h} \right) + \left( \frac{g(x+h) - g(x)}{h} \right) \right] \\ &= \lim\limits_{h \to 0} \left( \frac{f(x+h) - f(x) }{h} \right) + \lim\limits_{h \to 0} \left( \frac{g(x+h) - g(x)}{h} \right) \\ &= f'(x) + g'(x). \end{aligned} \]
Examples:
For any nonzero real number \(n\), if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
Proof (for positive integer \(n\)): \[ \begin{aligned} \frac{d}{dx} f(x) &= \lim\limits_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim\limits_{h \to 0} \frac{(x+h)^n - x^n}{h} \\ &= \lim\limits_{h \to 0} \frac{\left( x^n + n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + h^n \right) - x^n}{h} \\ &= \lim\limits_{h \to 0} \frac{n x^{n-1} h + \binom{n}{2} x^{n-2} h^2 + \dots + h^n}{h} \\ &= \lim\limits_{h \to 0} \left( n x^{n-1} + \binom{n}{2} x^{n-2} h + \dots + h^{n-1} \right) \\ &= n x^{n-1}. \end{aligned} \]
Note: The proof there uses the Binomial Theorem: \[ (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}. \]
Examples:
If \(f\) and \(g\) are differentiable functions, then their product \(P(x) = (f \cdot g)(x) = f(x) \cdot g(x)\) is also a differentiable function, and \[ P'(x) = f(x) g'(x) + g(x)f'(x). \]
I will leave out the proof here, but I can post it later. When we get to implicit differentiation and logarithms, it will become a lot easier to prove.
Examples:
If \(f\) and \(g\) are differentiable functions, then their quotient \(Q(x) = \frac{f(x)}{g(x)}\) is also a differentiable function for all \(x\) where \(g(x) \ne 0\), where \[ Q'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{\left(g(x)\right)^2}. \]
I like to remember this as “lo d-hi minus hi d-lo over lo squared”.
I will leave out the proof here, but I can post it later. When we get to implicit differentiation and logarithms, it will become a lot easier to prove.
Examples: