Question

How do you differentiate `f(x)=cotx/(1-sinx)`?

Answer 1

`d/(dx) ((cotx)/(1-sinx)) = color(blue)(((cotx)(cosx) + (-csc^2x)(1-sinx))/((1-sinx)^2)`

Explanation:

We can use the quotient rule:

`d/(dx) (u/v) = (v(du)/(dx) - u(dv)/(dx))/(v^2)`

where

`u = cotx`

and

`v = 1-sinx`

`(-cotx(d/(dx)(1-sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)`

`(-cotx(d/(dx)(1)-d/(dx)(sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)`

The derivative of `1` is `0`:

`(-cotx(-d/(dx)(sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)`

`(cotx(d/(dx)(sinx)) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)`

The derivative of `sinx` is `cosx`:

`((cotx)(cosx) + (d/(dx)(cotx))(1-sinx))/((1-sinx)^2)`

The derivative of `cotx` is `-csc^2x`:

`color(blue)(((cotx)(cosx) + (-csc^2x)(1-sinx))/((1-sinx)^2)`

or

`color(blue)(((cotx)(cosx))/((1-sinx)^2) - (csc^2x)/(1-sinx)`

Answer 2

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

Explanation:

In order to derive `cot(x)/(1-sin(x))`, you need to apply the quotient rule first.

The quotient rule states this:
`(f'(x)g(x) - g'(x)f(x))/g(x)^2`

Let's assume `(1-sin(x))` is `g(x)`.

Now, the `f(x)` that we will use for the quotient rule is not the equation they give us. It is the numerator, which is `cot(x)`. So let's plug in our newly found `f(x)` and `g(x)`, which is:

`(d/dxcot(x)(1-sin(x))-d/dx(1-sin(x))(cot(x)))/(1-sin(x))^2`

Now, the derivative of `cot(x)` is `-csc^2(x)` and the derivative of `(1-sin(x))` is `-cos(x)`.

After finding the derivatives of `f(x)` and `g(x)`, we rewrite our previous equation:

`(-csc^2(x)(1-sin(x))-(-cos(x))(cot(x)))/(1-sin(x))^2`

Now we split the numerator into two parts: `(cot(x)cos(x))/(1-sin(x))^2-(csc^2(x)(1-sin(x)))/(1-sin(x))^2`

For `(csc^2(x)(1-sin(x)))/(1-sin(x))^2` we can cancel out one `(1-sin(x))` from the numerator and denominator, so this simplifies to: `(csc^2(x))/(1-sin(x))`

So the final answer is: `f'(x) = (cos(x)cot(x))/(1-sin(x))^2-csc^2(x)/(1-sin(x))`