Question

How do you differentiate `y=e^x/x^7`?

Answer 1

`y^'=(e^x(x-7))/x^8`

Explanation:

The easiest way, for me, is to first write this not as a quotient:

`y=e^x/x^7=e^x x^-7`

From here, use the product rule, which states that if `y=f(x)g(x)`, then `y^'=f^'(x)g(x)+f(x)g^'(x)`.

So here, we see that `f(x)=e^x`, so `f^'(x)=e^x` as well, and `g(x)=x^-7`, so `g^'(x)=-7x^-8`.

Thus:

`y^'=e^x x^-7+e^x(-7x^-8)`

Simplifying:

`y^'=e^x/x^7-(7e^x)/x^8`

Common denominator:

`y^'=(xe^x-7e^x)/x^8`

`y^'=(e^x(x-7))/x^8`

Note that this can also be done with the quotient rule, which states that if `y=f(x)/g(x)` then `y^'=(f^'(x)g(x)-f(x)g^'(x))/(g(x))^2`.

So, in this case `f(x)=e^x` so again `f^'(x)=e^x`, but `g(x)=x^7` so `g^'(x)=7x^6`.

Thus:

`y^'=(e^x x^7-e^x(7x^6))/(x^7)^2=(e^x x^6(x-7))/x^14=(e^x(x-7))/x^8`