Differentiate a polynomial from first principles

Primary Tool

Math Workspace (CAS)

CAS is useful here because it keeps the algebraic expansion, the simplification, and the limiting step in one transcript.

Problem

Find the derivative of $f (x) = x^{2} + 3 x$ from the definition $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ .

This is a standard exercise because it shows that the derivative formula comes from algebraic cancellation, not from a rule you memorize first.

What to watch

When you substitute $x + h$ , the numerator looks messy on purpose. The cancellation is what reveals the derivative.

The limit is easy only after the numerator is simplified to $2 x + h + 3$ . Sending $h$ to $0$ too early loses the structure of the problem.

Step-by-step walkthrough

The goal is to make the limit readable by turning the difference quotient into a simpler algebraic expression.

1Start from $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ and substitute $f (x) = x^{2} + 3 x$ .
2Expand $f (x + h)$ as $(x + h)^{2} + 3 (x + h) = x^{2} + 2 x h + h^{2} + 3 x + 3 h$ .
3Subtract $f (x) = x^{2} + 3 x$ so the numerator becomes $2 x h + h^{2} + 3 h$ .
4Factor out $h$ to get $\frac{h ( 2 x + h + 3 )}{h}$ , then cancel the common factor for $h \neq = 0$ .
5Take the limit of the simplified expression $2 x + h + 3$ as $h \to 0$ to obtain $2 x + 3$ .

Common Pitfall

Do not substitute $h = 0$ into the original quotient. First simplify the expression, then take the limit of the simplified form as $h$ approaches $0$ .

Try a Variation

Try the same method on $f (x) = x^{3}$ . Which cancellation pattern repeats, and where does the extra power show up?

Keep moving through the cluster

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ExplanationCalculus2D Graphing

What a derivative means geometrically

The derivative is the slope of the tangent line, but that phrase only becomes useful when you compare the curve to nearby secant lines and local linear approximations.

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ProofReal AnalysisProof Builder

Epsilon-delta proof that $lim_{x \to 2} x^{2} = 4$

This proof is common because it shows the standard move in early analysis: factor the expression, then bound the extra factor by forcing $x$ into a smaller interval first.

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