## Calculus: Early Transcendentals 8th Edition

$f'(x)=8-10x$
$f(x)=4+8x-5x^{2}$ Derivative using the definition: $f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$ To find $f(x+h)$, wherever you see $x$ in the function, plug in $x+h$ $f(x+h)=4+8(x+h)-5(x+h)^{2}=...$ $...=4+8x+8h-5(x^{2}+2xh+h^{2})=...$ $...=4+8x+8h-5x^{2}-10xh-5h^{2}$ Let's plug in the components of the formula $f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}=\lim\limits_{h \to 0}\dfrac{4+8x+8h-5x^{2}-10xh-5h^{2}-4-8x+5x^{2}}{h}=...$ $...=\lim\limits_{h \to 0}\dfrac{8h-10xh-5h^{2}}{h}$ Take out common factor $h$ $...=\lim\limits_{h \to 0}\dfrac{h(8-10x-5h)}{h}=\lim\limits_{h \to 0}8-10x-5h=...$ Apply direct substitution to evaluate the limit: $f'(x)=8-10x-5(0)=8-10x$ $f'(x)=8-10x$