Derivative of Square Function/Proof 1

Theorem

Let $f: \R \to \R$ be the square function:

$\forall x \in \R: \map f x = x^2$

Then the derivative of $f$ is given by:

$\map {f'} x = 2 x$

$\ds \map {f'} x$

$=$

$\ds \lim_{h \mathop \to 0} \frac {\map f {x + h} - \map f x} h$

Definition of Derivative of Real Function at Point

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {\paren {x + h}^2 - x^2} h$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {x^2 + 2 x h + h^2 - x^2} h$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {2 x h + h^2} h$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} 2 x + h$

$\ds $

$=$

$\ds 2 x$

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Standard Differential Coefficients
1937: Eric Temple Bell: Men of Mathematics ... (previous) ... (next): Chapter $\text{VI}$: On the Seashore
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): differentiation
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): differentiation
2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $8$: The System of the World: Calculus