derivative of ((1+x)*e^x)
1) Let's differentiate: ddx((1 + x)·ex)
We differentiate u1 = (1 + x) and v2 = ex using: ddx(u1·v2) = ddx(u1)·v2 + u1·ddx(v2)
2) Let's differentiate u1: ddx(1 + x)
ⓘ The derivative of a sum is the sum of the derivatives: ddx(1 + x) = 0 + 1
0 + 1
ⓘ Adding 0 changes nothing: 0 + 1 = 1
1
3) Let's differentiate v2: ddx(ex)
ⓘ The derivative of exp(x) is exp(x): ddx(ex) = ex
ex
4) We evaluate the resulting derivative (u1·v2)': 1·ex + (1 + x)·ex
ex + (1 + x)·ex
ⓘ Expand by multiplying each term inside the parentheses: (1 + x)·ex = ex + xex
ex + ex + xex
ⓘ Like terms are grouped by adding their coefficients: ex + ex = 2ex
2ex + xex
5) You can then resume the derivation:
ⅰ ddx((1 + x)·ex) = 2ex + xex
2ex + xex
exx + 2ex
ex·(x + 2)
▶Derivative:   ex·(x + 2)