How do you differentiate #y=1/sinx+1/cosx#?

2 Answers
Oct 31, 2016

# dy/dx = -cotx cscx + tanx secx #

Explanation:

You should learn that #1/sinx=cscx# and #1/cosx=secx#
And, #d/dxcscx=-csc x cot x# and #d/dxsecx=sec x tan x#

These resul;ts can be obtained using the quotient rule;
# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #;

# y = 1/sinx + 1/cosx #

# :. dy/dx = { ( (sinx)(d/dx1) - (1)(d/dxsinx) ) / (sinx)^2 } + { ( (cos)(d/dx1) - (1)(d/dxcosx) ) / (cosx)^2 } #

# :. dy/dx = { ( 0 - (1)(cosx) ) / (sinx)^2 } + { ( 0 - (1)(-sinx) ) / (cosx)^2 } #

# :. dy/dx = -cosx/ (sinx)^2 + sinx / (cosx)^2 #
# :. dy/dx = -cosx/sinx 1/sinx + sinx/cosx 1/cosx #
# :. dy/dx = -cotx cscx + tanx secx #

Oct 31, 2016

#y=1/sinx+1/cosx#

#y=cscx+secx#

#(dy)/(dx)=-cotxcscx+secxtanx#

To find out how to differentiate #cscx# using implicit differentiation watch this video:

To find out how to differentiate #secx# using implicit differentiation watch this video: