How do you find the derivative of #2x^2+x-1#?

1 Answer
Jul 16, 2015

#(d(2x^2+x-1))/(dx) = 4x+1#

Explanation:

The derivative of #f(x)= p(x)+q(x)+r(x)#:
#color(white)("XXXX")##(d(f(x)))/(dx) = (d(p(x)))/(dx) + (d(q(x)))/(dx) + (d(r(x)))/(dx)#
That is, the derivative of a sum of (sub)functions is the sum of the derivatives of the (sub)functions.

The derivative of #ax^b#
#color(white)("XXXX")##(d(ax^b))/(dx) = bax^(b-1)#

Using
#color(white)("XXXX")##p(x) = 2x^2#
#color(white)("XXXX")##color(white)("XXXX")##rarr (d(p(x)))/(dx) = 4x^1#

#color(white)("XXXX")##q(x) = x (= 1x^1)#
#color(white)("XXXX")##color(white)("XXXX")##rarr (d(q(x)))/(dx) = (1)1x^0 = 1#

#color(white)("XXXX")##r(x) = -1 (= (-1)x^0)#
#color(white)("XXXX")##color(white)("XXXX")##rarr (d(r(x)))/(dx) = (0)(-1)x^(-1) = 0#

#(d(f(x)))/(dx) = 4x + 1 (+0)#
#color(white)("XXXX")#when #f(x) = 2x^2+x-1#

(I recognize that this is a long explanation to a simple problem, but I wanted to try to cover any possible problems in following the solution).