Question

How do you differentiate `f(x)= cosx/ (sinx)` twice using the quotient rule?

Answer 1

`f'(x)=-csc^2x,f''(x)=2csc^2xcotx`

Explanation:

Through the quotient rule:

`f'(x)=(sinxd/dx[cosx]-cosxd/dx[sinx])/sin^2x`

Recall that:

`d/dx[cosx]=-sinx`

`d/dx[sinx]=cosx`

`f'(x)=(-sin^2x-cos^2x)/sin^2x`

`f'(x)=-(sin^2x+cos^2x)/sin^2x`

`f'(x)=-1/sin^2x`

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

In order to find the second derivative, if we want to continue using the quotient rule, use:

`f'(x)=(-1)/sin^2x`

`f''(x)=(sin^2xd/dx[-1]-(-1)d/dx[sin^2x])/sin^4x`

Again, find each derivative:

`d/dx[-1]=0`

Chain rule coming up:

`d/dx[sin^2x]=2sinxd/dx[sinx]=2sinxcosx`

Plug back in:

`f''(x)=(2sinxcosx)/sin^4x`

`f''(x)=(2cosx)/sin^3x`

`f''(x)=2(1/sin^2x)(cosx/sinx)`

`f''(x)=2csc^2xcotx`