The derivative of f(x)= p(x)+q(x)+r(x)f(x)=p(x)+q(x)+r(x):
color(white)("XXXX")XXXX(d(f(x)))/(dx) = (d(p(x)))/(dx) + (d(q(x)))/(dx) + (d(r(x)))/(dx)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^baxb
color(white)("XXXX")XXXX(d(ax^b))/(dx) = bax^(b-1)d(axb)dx=baxb−1
Using
color(white)("XXXX")XXXXp(x) = 2x^2p(x)=2x2
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXrarr (d(p(x)))/(dx) = 4x^1→d(p(x))dx=4x1
color(white)("XXXX")XXXXq(x) = x (= 1x^1)q(x)=x(=1x1)
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXrarr (d(q(x)))/(dx) = (1)1x^0 = 1→d(q(x))dx=(1)1x0=1
color(white)("XXXX")XXXXr(x) = -1 (= (-1)x^0)r(x)=−1(=(−1)x0)
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXXrarr (d(r(x)))/(dx) = (0)(-1)x^(-1) = 0→d(r(x))dx=(0)(−1)x−1=0
(d(f(x)))/(dx) = 4x + 1 (+0)d(f(x))dx=4x+1(+0)
color(white)("XXXX")XXXXwhen f(x) = 2x^2+x-1f(x)=2x2+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).
