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

Question

Christopher Allers · Accepted Answer

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

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#

Although this is a lengthy explanation for a straightforward issue, I wanted to make sure that any potential issues with applying the solution were covered.

Paisley Johnson · Answer

undefined To find the derivative of $2x^2 + x - 1$, you apply the power rule for differentiation. The derivative of $x^n$ with respect to $x$ is $nx^{n-1}$. So, the derivative of $2x^2$ is $4x$, the derivative of $x$ is $1$, and the derivative of the constant term $-1$ is $0$. Therefore, the derivative of $2x^2 + x - 1$ is $4x + 1$.