Question

How do you find the first and second derivatives of `f(x)=cos(x)/x^8` using the quotient rule?

Answer 1

First derivative: `-\frac{x\sin(x)+8\cos(x)}{x^9}`
Second derivative: `\frac{7\sin(x)-x\cos(x)}{x^9}-\frac{9(\sin(x)-8\cos(x))}{x^10}`

Explanation:

The quotient `\frac[f(x)}{g(x)}` of two derivable functions `f(x)` and `g(x)` is derivable for all `x` in which `g(x)\ne 0`. And its derivative is equal to:

`frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{(g(x))^2}`

So, going back to the exercise we have:

`f(x)=\sin(x)`
`g(x)=x^8`

Thus:

`frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{x^8\cdot (-\sin(x)) - \cos(x)\cdot 8x^7}{(x^8)^2}`

`frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^8\sin(x)-8x^7\cos(x)}{x^16}`

Factoring:

`frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^7(x\sin(x)+8\cos(x))}{x^16}`

Reducing `frac{x^7}{x^16}` we finally get:

`frac{d}{dx}(\frac[f(x)}{g(x)})=-\frac{x\sin(x)+8\cos(x)}{x^9}`

For the second derivative your functions `f(x)` and `g(x)` will be:

`f(x)=-(x\sin(x)+8\cos(x))`
`g(x)=x^9`

Try doing it!