How do you use the definition of a derivative to find the derivative of #f(x)=x^2-3x-1#?

Question

Cameron Alexander · Accepted Answer

#f'(x)=2x-3# The definition of the derivative is: #f'(x)=lim_(h->0)(f(x+h)-f(x))/h# Applying this to the problem: #f'(x)=lim_(h->0)((x+h)^2-3(x+h)-1-(x^2-3x-1))/h# #f'(x)=lim_(h->0)(x^2+2xh+h^2-3x-3h-1-x^2+3x+1)/h# #f'(x)=lim_(h->0)(2xh+h^2-3h)/h# #f'(x)=lim_(h->0)(h(2x+h-3))/h# #f'(x)=lim_(h->0)2x+h-3# #f'(x)=2x+(0)-3=2x-3#

Charles Arocha · Answer

#(df(x))/dx=2x-3#

Definition of derivative is given by

#(df(x))/dx=Lt_(deltax->0)(f(x+deltax)-f(x))/(deltax)#

As #f(x)=x^2-3x-1#, hence

#f(x+deltax)-f(x)=(x+deltax)^2-3(x+deltax)-1-(x^2-3x-1)#

= #(x^2+2xdeltax+(deltax)^2-x^2)-3(x+deltax-x)-1+1#

= #(2xdeltax+(deltax)^2)-3(deltax)# and

#f((x+deltax)-f(x))/(deltax)=2x-3+deltax#

Hence, #Lt_(deltax->0)(f(x+deltax)-f(x))/(deltax)#

= #Lt_(deltax->0)(2x-3+deltax)# i.e.

#(df(x))/dx=2x-3#

Evelyn Agard · Answer

undefined To find the derivative of $ f(x) = x^2 - 3x - 1 $ using the definition of a derivative, you would apply the limit definition of the derivative, which is:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute $ f(x) = x^2 - 3x - 1 $ into the formula and then simplify the expression.