Question

How do you find the derivative of `w=1/sinz`?

Answer 1

`(dw)/(dz)=-cos(z)/(sin^2(z))`

Explanation:

For the general case, the derivative quotient rule tell us:
`color(white)("XXX")(d (f_x))/(d (g_x)) =((df_x)/(dx) * g_x - (dg_x)/(dx) * f_x)/(g_x^2)`

Taking `f(z) = 1` and `g(z)=sin(z)`
(so `w_z=(f_z)/(g_z)`)
and
remembering that
`color(white)("XXX")(d sin(z))/(dz)=cos(z)`
we have
`color(white)("XXX")(dw_z)/(dz)=(0 * sin(z) - cos(z) * 1)/(sin^2(z))`

`color(white)("XXXXX")=(-cos(z))/(sin^2(z))`

Answer 2

`(dw)/dz = -csc(z) cot(z) = (-cosz)/(sin^2z)`

Explanation:

The answer below is also valid, but here is a shortcut if you can remember the identity:

`w = 1/sinz = cscz`

`therefore (dw)/dz = d/dz csc z = -csc(z) cot(z)`

So the answer can be written as `-csc(z) cot(z)` or `(-cosz)/(sin^2z)` since they are equivalent forms.

Final Answer

---

This is a common basic identity which can be quickly memorized along with d/dx of sine, cosine, tangent, etc. in order to save time in the future:

`d/dx csc(x) = -csc(x) cot(x)`