Expand d f d x {\displaystyle {\frac {df}{dx}}} as a Maclaurin series by using the geometric series 1 1 + y = 1 1 − ( − y ) = 1 − y + y 2 − y 3 + ⋯ = ∑ n = 0 ∞ ( − y ) n . {\displaystyle {\frac {1}{1+y}}={\frac {1}{1-(-y)}}=1-y+y^{2}-y^{3}+\cdots =\sum _{n=0}^{\infty }(-y)^{n}.}
as a Maclaurin series by using the geometric series 
