## Calculus: Early Transcendentals 8th Edition

$f'(x)=m$
$f(x)=mx+b$ Derivative of a function using the definition: $f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$ To find $f(x+h)$, wherever you find $x$ in the function, substitute $x+h$ $f(x+h)=m(x+h)+b=mx+mh+b$ Let's plug in the components of the formula: $f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h \to 0}\dfrac{mx+mh+b-mx-b}{h}=\lim\limits_{h \to 0}\dfrac{mh}{h}=\lim\limits_{h \to 0}\ m=m$ $f'(x)=m$