How do you find the first and second derivative of #x^2-2x-3#?

1 Answer
May 12, 2016

If #f(x)=x^2-2x-3#
#color(white)("XXX")f'(x)=2x-2# and
#color(white)("XXX")f''(x)=2#

Explanation:

The first derivative:
#f'(x)=(df(x))/(dx)=(dcolor(red)(x^2))/(dx)-(dcolor(blue)(2x))/(dx)-(dcolor(green)(3))/(dx)#

#color(white)("XXXXXXXXX")=color(red)(2x^1)-color(blue)(2x^0)-color(green)(0)#

#color(white)("XXXXXXXXX")=color(red)(2x)-color(blue)(2)#

The second derivative
The second derivative is just the derivative of the first derivative.
#f''(x)=(df'(x))/(dx) = (dcolor(red)(2x^1))/(dx)-(dcolor(blue)(2x^0))/(dx)#

#color(white)("XXXXXXXXXX")=color(red)(2x^0)-color(blue)(0)#

#color(white)("XXXXXXXXXX")=color(red)(2)#