Question

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

Answer 1

`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)|)))`