How do you differentiate #p(y) = (y^-1 + y^-2)(5y^-3 + 9y^-4)# using the product rule?

1 Answer
Jul 1, 2017

Answer:

#(dp(y))/(dy)=-20y^(-5)-70y^(-6)-54y^(-7)#

Explanation:

According to product rule derivative of #p(y)=f(y)g(y)# is

#(dp(y))/(dy)=(df)/(dy)xxg(y)+f(y)xx(dg)/(dy)#

Here we have #f(y)=y^(-1)+y^(-2)# and #(df)/(dy)=-y^(-2)-2y^(-3)#

and #g(y)=5y^(-3)+9y^(-4)# and #(dg)/(dy)=-15y^(-4)-36y^(-5)#

Therefore

#(dp(y))/(dy)=(-y^(-2)-2y^(-3))xx(5y^(-3)+9y^(-4))+(y^(-1)+y^(-2))xx(-15y^(-4)-36y^(-5))#

= #-5y^(-5)-9y^(-6)-10y^(-6)-18y^(-7)-15y^(-5)-36y^(-6)-15y^(-6)-36y^(-7)#

= #-20y^(-5)-70y^(-6)-54y^(-7)#