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For f(x)=5x−x2 , use the definition of a derivative to find f′(x)

User Ilkin
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\bf f(x)=5x-x^2\qquad \qquad \lim\limits_(h\to 0)~\cfrac{f(x+h)-f(x)}{h} \\\\[-0.35em] ~\dotfill\\\\ \lim\limits_(h\to 0)~\cfrac{[5(x+h)-(x+h)^2]~~-~~[5x-x^2]}{h} \\\\\\ \lim\limits_(h\to 0)~\cfrac{[5x+5h-(x^2+2xh+h^2)]~~-~~5x+x^2}{h}


\bf \lim\limits_(h\to 0)~\cfrac{\begin{matrix} 5x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}+5h~~\begin{matrix} -x^2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}-2xh-h^2~~-~~\begin{matrix} 5x +x^2\\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}{h}\implies \lim\limits_(h\to 0)~\cfrac{5h-2xh-h^2}{h}


\bf \lim\limits_(h\to 0)~\cfrac{\begin{matrix} h \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} ~~(5-2x-h)}{\begin{matrix} h \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} }\implies \lim\limits_(h\to 0)~5-2x-0\implies \lim\limits_(h\to 0)~5-2x

User Hamid Mayeli
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