How do you differentiate #y(m)=(e^mroot3(m))/(m^2+3)# using the quotient rule?

1 Answer
Apr 13, 2017

#y'(m)=(e^m(3m^3-5m^2+9m+3))/(3m^(2/3)(m^2+3)^2)#

Explanation:

The quotient rule states that where:

#y(m)=(f(m))/(g(m))#

then its derivative is

#y'(m)=(f'(m)g(m)-f(m)g'(m))/(g(m))^2#

For #y(m)=(e^mroot3m)/(m^2+3)# we see that:

#{(f(m)=e^mroot3m),(g(m)=m^2+3):}#

To find the derivative of #f#, we will need to use the product rule. A quick review of the product rule is that the derivative of #r(m)s(m)# is #r'(m)s(m)+r(m)s'(m)#.

Rewriting #f# as #f(m)=e^mm^(1/3)#, we see that:

#f'(m)=(d/(dm)e^m)m^(1/3)+e^m(d/(dm)m^(1/3))#

#f'(m)=e^mm^(1/3)+e^m(1/3m^(-2/3))#

Simplifying by multiplying by #(3m^(2/3))/(3m^(2/3))#:

#f'(m)=(3me^m+e^m)/(3m^(2/3))=(e^m(3m+1))/(3m^(2/3))#

And:

#g'(m)=2m#

Then, through the quotient rule:

#y'(m)=((e^m(3m+1))/(3m^(2/3))(m^2+3)-(e^mm^(1/3)(2m)))/(m^2+3)^2#

Again multiplying through by #3m^(2/3)# in the numerator and denominator:

#y'(m)=(e^m(3m+1)(m^2+3)-6m^2e^m)/(3m^(2/3)(m^2+3)^2)#

#y'(m)=(e^m[(3m+1)(m^2+3)-6m^2])/(3m^(2/3)(m^2+3)^2)#

#y'(m)=(e^m(3m^3-5m^2+9m+3))/(3m^(2/3)(m^2+3)^2)#