How do you differentiate #h(y)=1/(y^3+2y+1)# using the quotient rule?

1 Answer
mason m
Aug 25, 2016

#h'(y)=(-3y^2-2)/(y^3+2y+1)^2#

Explanation:

The quotient rule states that for some function in the form #h(y)=f(y)/g(y)#, the function's derivative equals #h'(y)=(f'(y)g(y)-f(y)g'(y))/(g(y))^2#.

Here, for #h(y)=1/(y^3+2y+1)#, our two functions being divided are:

#{(f(y)=1),(g(y)=y^3+2y+1):}#

We will need to know their derivatives in a moment:

#{(f'(y)=0),(g'(y)=3y^2+2):}#

Plugging these into the quotient rule formula gives us:

#h'(y)=((0)(y^3+2y+1)-(1)(3y^2+2))/(y^3+2y+1)^2#

Simplifying:

#h'(y)=(-3y^2-2)/(y^3+2y+1)^2#

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