Derivatives: Learning Objectives 1, 2, 3
Question 1: State the limit definition of the derivative.
The definition involves the limit of the slope of a secant line as the distance between the points approaches zero.
$$ f'(x) = \lim_{h\to 0}\frac{f(x + h) - f(x)}{h} $$
The derivative of a function $f$ at a number $x$, denoted by $f'(x)$, is given by the following limit, provided the limit exists:
$$ f'(x) = \lim_{h\to 0}\frac{f(x + h) - f(x)}{h} $$
Question 2: Please provide a geometric interpretation for the limit definition of the derivative.
The expression $\frac{f(x+h) - f(x)}{h}$ represents the slope of a line. What kind of line is it, and what happens as $h \to 0$?
The derivative $f'(x)$ is the slope of the tangent line to the curve $y=f(x)$ at the point $(x, f(x))$.
The expression $\frac{f(x+h) - f(x)}{(x+h)-x} = \frac{f(x+h) - f(x)}{h}$ is the slope of the secant line connecting the points $(x, f(x))$ and $(x+h, f(x+h))$ on the curve $y=f(x)$.
As we take the limit $h \to 0$, the point $(x+h, f(x+h))$ gets closer and closer to the point $(x, f(x))$. The secant line pivots and approaches a limiting position, which is the tangent line to the curve at the point $(x, f(x))$.
Therefore, the limit of the slopes of the secant lines is the slope of the tangent line. This means the derivative represents the instantaneous rate of change of the function at a specific point.
As we take the limit $h \to 0$, the point $(x+h, f(x+h))$ gets closer and closer to the point $(x, f(x))$. The secant line pivots and approaches a limiting position, which is the tangent line to the curve at the point $(x, f(x))$.
Therefore, the limit of the slopes of the secant lines is the slope of the tangent line. This means the derivative represents the instantaneous rate of change of the function at a specific point.
Question 3: Evaluate the derivative for the following polynomial using the limit definition of the derivative.
$$f(x) = x^2 + 2x - 5$$
Substitute $f(x+h)$ and $f(x)$ into the limit definition, expand the terms, simplify the numerator, and then evaluate the limit.
$f'(x) = 2x+2$
\begin{align*}
f'(x) & = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \\
& = \lim_{h\to 0} \frac{[(x+h)^2 + 2(x+h) - 5] - [x^2 + 2x - 5]}{h} \\
& = \lim_{h\to 0} \frac{(x^2 + 2xh + h^2 + 2x + 2h - 5) - x^2 - 2x + 5}{h} \\
& = \lim_{h\to 0} \frac{2xh + h^2 + 2h}{h} \\
& = \lim_{h\to 0} \frac{h(2x + h + 2)}{h} \\
& = \lim_{h\to 0} (2x + h + 2)\\
& = 2x + 0 + 2 = 2x + 2
\end{align*}
Question 4: A function $f(x)$ has derivative $f'(x)$. Show (using the limit definition of the derivative) that for $a \in \mathbb{R}$, the derivative of $g(x) = a \cdot f(x)$ is $g'(x) = a \cdot f'(x)$.
Apply the limit definition to $g(x)$. You will be able to factor out the constant $a$ from the limit.
The proof uses the limit definition of the derivative for $g(x)$, followed by applying the Constant Multiple Law for limits.
\begin{align*}
g'(x) & = \lim_{h\to 0} \frac{g(x+h) - g(x)}{h} \\
& = \lim_{h\to 0} \frac{a \cdot f(x+h) - a \cdot f(x)}{h} \quad (\text{by definition of } g(x)) \\
& = \lim_{h\to 0} a \cdot \frac{f(x+h) - f(x)}{h} \quad (\text{factor out constant } a) \\
& = a \cdot \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \quad (\text{by Constant Multiple Law for limits}) \\
& = a \cdot f'(x) \quad (\text{by definition of } f'(x))
\end{align*}
Question 5: Functions $f(x)$ and $g(x)$ have derivatives $f'(x)$ and $g'(x)$, respectively. Show (using the limit definition of the derivative) a function $j(x) = f(x) + g(x)$ has derivative $j'(x) = f'(x) + g'(x)$.
Apply the limit definition to $j(x)$. Rearrange the terms in the numerator and use the Sum Law for limits.
The proof uses the limit definition for $j(x)$ and applies the Sum Law for limits to separate the expression into the limit definitions for $f'(x)$ and $g'(x)$.
\begin{align*}
j'(x) & = \lim_{h\to 0}\frac{j(x+h) - j(x)}{h} \\
& = \lim_{h\to 0}\frac{[f(x+h) + g(x+h)] - [f(x) + g(x)]}{h} \quad (\text{by definition of } j(x)) \\
& = \lim_{h\to 0}\frac{f(x+h) - f(x) + g(x+h) - g(x)}{h} \quad (\text{rearrange terms}) \\
& = \lim_{h\to 0} \left( \frac{f(x+h)-f(x)}{h} + \frac{g(x+h) -g(x)}{h} \right) \quad (\text{split fraction}) \\
& = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} + \lim_{h\to 0}\frac{g(x+h) -g(x)}{h} \quad (\text{by Sum Law for limits}) \\
& = f'(x) + g'(x) \quad (\text{by definition of } f'(x) \text{ and } g'(x))
\end{align*}
Question 6: Use the limit definition of the derivative to show that the derivative of $f(x) = x^n$ is $f'(x) = nx^{n-1}$.
Hint: Use the binomial theorem to evaluate $(x+h)^n$.
Hint: Use the binomial theorem to evaluate $(x+h)^n$.
The Binomial Theorem states that $(a+b)^n = \sum_{i=0}^n \binom{n}{i} a^{n-i}b^i = a^n + na^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n$.
The derivative is $nx^{n-1}$. The proof involves expanding $(x+h)^n$, canceling terms, dividing by $h$, and taking the limit, where only the second term of the expansion remains.
We start with the limit definition:
\[ f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{(x+h)^n - x^n}{h} \]
Using the Binomial Theorem, we expand $(x+h)^n$:
\[ (x+h)^n = \binom{n}{0}x^n h^0 + \binom{n}{1}x^{n-1}h^1 + \binom{n}{2}x^{n-2}h^2 + \dots + \binom{n}{n}x^0h^n \]
\[ = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n \]
Substitute this back into the limit:
\[ f'(x) = \lim_{h\to 0} \frac{(x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots) - x^n}{h} \]
The $x^n$ terms cancel:
\[ = \lim_{h\to 0} \frac{nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots}{h} \]
Divide every term in the numerator by $h$:
\[ = \lim_{h\to 0} \left( nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots \right) \]
As $h \to 0$, every term except the first one goes to zero because they all contain a factor of $h$.
\[ = nx^{n-1} + 0 + 0 + \dots = nx^{n-1} \]