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

2 Answers
Mar 7, 2017

#(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))#

Mar 19, 2017

#(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)#