Using the limit definition, how do you find the derivative of #y =3/(x+1)#?
1 Answer
May 10, 2016
Explanation:
The limit definition of a derivative states that
#f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h#
Substituting
#f'(x)=lim_(hrarr0)(3/((x+h)+1)-3/(x+1))/h#
From this point on, you want to expand and simplify.
#f'(x)=lim_(hrarr0)(3/(x+h+1)((x+1)/(x+1))-3/(x+1)((x+h+1)/(x+h+1)))/h#
#f'(x)=lim_(hrarr0)((3(x+1))/(x^2+xh+x+x+h+1)-(3(x+h+1))/(x^2+xh+x+x+h+1))/h#
#f'(x)=lim_(hrarr0)((3x+3)/(x^2+2x+xh+h+1)-(3x+3h+3)/(x^2+2x+xh+h+1))/h#
#f'(x)=lim_(hrarr0)((-3h)/(x^2+2x+xh+h+1))/h#
#f'(x)=lim_(hrarr0)(-3h)/(x^2+2x+xh+h+1)*1/h#
#f'(x)=lim_(hrarr0)-3/(x^2+2x+xh+h+1)#
Plugging in
#f'(x)=-3/(x^2+2x+x(0)+(0)+1)#
#f'(x)=-3/(x^2+2x+1)#
#color(green)(|bar(ul(color(white)(a/a)color(black)(f'(x)=-3/((x+1)^2)color(white)(a/a)|)))#