Question

How do you differentiate `f(x)= ( x + 1 )/ ( tan x )` using the quotient rule?

Answer 1

`f'(x) = cotx-xcsc^2x-csc^2x`

Explanation:

The quotient rule states that:
`d/dx(u/v) = (u'v-uv')/v^2`
Where `u` and `v` are functions of `x`.

We use this whenever we need to find the derivative of one function divided by another, which is the case here. We can say let `u=x+1` and `v=tanx`; finding the derivatives of each of these individually:
`u'=1`
`v'=sec^2x`

Substituting these into the quotient rule formula, we have:
`f'(x) = ((x+1)'(tanx)-(x+1)(tanx)')/(tanx)^2`
`f'(x) = (tanx-(x+1)(sec^2x))/(tan^2x)`

Now we move onto the algebra/trig, which tends to be the most difficult part:
`f'(x) = tanx/tan^2x-((x+1)(sec^2x))/(tan^2x)`
`f'(x) = cotx-((x+1)(1/cos^2x))/(sin^2x/cos^2x)`
`f'(x) = cotx-(x+1)(1/cos^2x)*(cos^2x/sin^2x)`
`f'(x) = cotx-(x+1)(csc^2x)`
`f'(x) = cotx-(xcsc^2x+csc^2x)`
`f'(x) = cotx-xcsc^2x-csc^2x`

We could do more simplifying here, but this result suffices.