Question

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

Answer 1

`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)`