How do you use cancellation to find the derivative of $f(x)=x^2$ from the limit definition?

Q: How do you use cancellation to find the derivative of $f(x)=x^2$ from the limit definition?

To find the derivative of $f(x)=x^2$, substitute into the difference quotient to get $\frac{(x+h)^2-x^2}{h}$. Expanding the numerator yields $2xh+h^2$. For nonzero $h$, you can cancel $h$ to obtain $2x+h$. Taking the limit as $h \to 0$ gives the derivative $2x$.

To find the derivative of

f(x)=x^2

, substitute into the difference quotient to get

\frac{(x+h)^2-x^2}{h}

. Expanding the numerator yields

2xh+h^2

. For nonzero

h

, you can cancel

h

to obtain

2x+h

. Taking the limit as

h \to 0

gives the derivative

2x

Conditions

$f(x)=x^2$ .
$h$ is nonzero during the cancellation step.
The limit is taken as $h \to 0$ after simplification.

Reasoning, step by step

Substitute $f(x+h)=(x+h)^2$ and $f(x)=x^2$ into the difference quotient.
Expand the numerator: $(x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2$ .
Factor out $h$ from the numerator: $h(2x+h)$ .
Cancel $h$ from the numerator and denominator, valid since $h \neq 0$ : $\frac{h(2x+h)}{h} = 2x+h$ .
Evaluate the limit: $\lim_{h\to0} (2x+h) = 2x$ .

Example

The script states: 'For $f(x)=x$ ², expanding the numerator gives 2xh+h². For nonzero h the quotient becomes $2x+h$ , whose limit is 2x.'

Common misconceptions

Attempting to cancel $h$ before expanding or simplifying the numerator.
Substituting $h=0$ into the unsimplified quotient, leading to division by zero.
Forgetting that cancellation is only valid for nonzero $h$ , though the limit considers the behavior as $h$ approaches zero.

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How do you use cancellation to find the derivative of $f(x)=x^2$ from the limit definition?

Conditions

Reasoning, step by step

Example

Common misconceptions

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Connected concepts

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Conditions

Reasoning, step by step

Example

Common misconceptions

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How do you use cancellation to find the derivative of $f(x)=x^2$ from the limit definition?