Derive the derivative of a square from first principles

$f(x+h)=(x+h)^2$. For $h\ne0$, $[f(x+h)-f(x)]/h=[x^2+2xh+h^2-x^2]/h=2x+h$. The source uses $\delta x$ in place of $h$ and explicitly advises expanding and simplifying.

$f^{\prime}(x)=\lim_{h\to0}(2x+h)=2x$, as required. The secant slope becomes the tangent slope in this limit. Do not set $h=0$ in the original quotient; cancellation is performed for nonzero $h$ before taking a limit. The source points to Appendix Examples A–C, PDF 17–20, for first-principles derivatives of $1/x$, $\sin x$ and $e^x$; those proofs are now included as Appendix Examples A–C below.