How do you use the definition of a derivative to find the derivative of #f(x) = 4 + 9x - x^2#?

1 Answer
May 4, 2016

#f'(x)=-2x+9#

Explanation:

The limit definition of a derivative states that

#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#

Substituting #f(x)=4+9x-x^2# into #f'(x)#,

#f'(x)=lim_(hrarr0)((4+9(x+h)-(x+h)^2)-(4+9x-x^2))/h#

From this point on, you want to expand and simplify.

#f'(x)=lim_(hrarr0)((4+9x+9h-(x^2+2xh+h^2))-(4+9x-x^2))/h#

#f'(x)=lim_(hrarr0)(4+9x+9h-x^2-2xh-h^2-4-9x+x^2)/h#

#f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)4color(blue)cancelcolor(black)(+9x)+9hcolor(teal)cancelcolor(black)(-x^2)-2xh-h^2color(red)cancelcolor(black)(-4)color(blue)cancelcolor(black)(-9x)color(teal)cancelcolor(black)(+x^2))/h#

#f'(x)=lim_(hrarr0)(9h-2xh-h^2)/h#

#f'(x)=lim_(hrarr0)(color(red)cancelcolor(black)h(9-2x-h))/color(red)cancelcolor(black)h#

#f'(x)=lim_(hrarr0)(9-2x-h)#

Plugging in #h=0#,

#f'(x)=(9-2x-0)#

#f'(x)=9-2x#

#color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=-2x+9)color(white)(a/a)|)))#