Question

How do you find the derivative `1/(x^7 + 2)`?

Answer 1

I would usr the Quotient Rule as:
`f'(x)=(-7x^6)/(x^7+2)^2`

Remember the Quotient Rule to derive a function `f(x)=(g(x))/(h(x))`
you get:
`f'(x)=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2`

Answer 2

To find the derivative of `f(x) = 1/(x^7 + 2)` you would use the quotient rule:

`f(x) = g(x)/(h(x))` then `f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x)^2)`

Thus:

let `g(x) = 1` so `g'(x) = 0`

let `h(x) = (x^7 + 2)` so `h'(x) = 7x^6`

Therefore we get that:

`f'(x) = (-7x^6)/(x^7 + 2)^2`

this shows us that `f(x)` is always decreasing, as its derivative is always negative

Answer 3

I would not use the quotient rule. (There's no difference in the answer, but this is how I think about this differentiation.)

I would use the chain rule:

`f(x)=1/(x^7+2) = (x^7+2)^-1`

`f'(x) = -1(x^7+2)^-2 * 7x^6`

`= (-7x^6)/(x^7+2)^2`