Question

How do you differentiate `f(x)=x^6/e^(x-6)` using the quotient rule?

Answer 1

`f'(x)=(x^5(6-x))/e^(x-6)`

Explanation:

`f'(x)=(u'v-uv')/v^2`

`u=x^6 => u'=6x^5`
`v=e^(x-6) => v'=e^(x-6)*(x-6)'=e^(x-6)*1=e^(x-6)`

`f'(x)=(6x^5e^(x-6)-x^6e^(x-6))/(e^(x-6))^2 = (x^5e^(x-6)(6-x))/(e^(x-6))^2`

`f'(x)=(x^5(6-x))/e^(x-6)`