Derivative of x² at 1 from the definition
Let $f(x) = x^2$. Using the definition of the derivative as a limit, find $f'(1)$.
Explanation of Method
The derivative of $f$ at a point $x_0$ is defined as the limit of the difference quotient, $f'(x_0) = \lim_{h\to0}\dfrac{f(x_0+h)-f(x_0)}{h}$, which measures the slope of secant lines through $(x_0, f(x_0))$ as the second point slides in toward $x_0$. For a polynomial like $f(x)=x^2$, expanding $f(x_0+h)$ turns the difference quotient into a simple algebraic expression in which $h$ can be factored out and cancelled, leaving an expression whose limit is found by direct substitution.
- Write down the definition of the derivative at $x_0=1$.
- Expand $f(1+h)$ and substitute both function values.
- Simplify the difference quotient by factoring out $h$.
- Take the limit as $h \to 0$.
Step 1 · Write down the definition
The derivative of $f$ at $x_0=1$ is defined as:
$$ f'(1) = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h}. $$
Step 2 · Substitute $f(x)=x^2$
Expanding $f(1+h) = (1+h)^2$:
$$ f(1+h) = 1+2h+h^2, \qquad f(1) = 1^2 = 1. $$
So the difference quotient becomes:
$$ f'(1) = \lim_{h\to0} \frac{(1+2h+h^2)-1}{h} = \lim_{h\to0}\frac{2h+h^2}{h}. $$
Step 3 · Factor and cancel $h$
Factoring $h$ out of the numerator:
$$ \frac{2h+h^2}{h} = \frac{h(2+h)}{h} = 2+h \qquad (h \neq 0). $$
Step 4 · Take the limit
Since the simplified expression $2+h$ is continuous, the limit is found by direct substitution:
$$ f'(1) = \lim_{h \to 0} (2+h) = 2. $$
Summary Table
| Quantity | Expression | Value |
|---|---|---|
| $f(1+h)-f(1)$ | $2h+h^2$ | — |
| Difference quotient | $2+h$ (after cancelling $h$) | — |
| $f'(1)$ | $\lim_{h\to0}(2+h)$ | $2$ |
Additional Notes
- The crucial algebraic step is factoring $h$ out of the numerator before taking the limit — direct substitution of $h=0$ into the original quotient $\frac{2h+h^2}{h}$ would give the indeterminate form $\frac00$, so the cancellation must happen first.
- This computation confirms the well-known power rule $\frac{d}{dx}(x^2) = 2x$ at the specific point $x=1$, since $2x$ evaluated at $x=1$ gives $2$.
- The same limit-definition approach works for any polynomial: expanding $f(x_0+h)$, subtracting $f(x_0)$, and dividing by $h$ always leaves a polynomial in $h$ with no constant term in the numerator, so an overall factor of $h$ always cancels cleanly.
Result
The derivative of $f(x)=x^2$ at $x_0=1$ is: