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What is the derivative of #(12/sinx) + (1/cotx)#?

Answer 1

#−12cosecx⋅cotx+sec^2x#

1/#sin x# = #cosec x# similarly #1/cotx# = #tanx# the question changes to #12cosecx+tanx# the derivative is #-12cosecx*cotx+sec^2x# by (u/v rule of differentiation) #d/dx# #(1/sinx)# = #1/sin^2x# #(-1*cosx)# = #-cosx/sinx# #1/sinx# =#-cosecx cotx# #d/dx# #tanx#= #d/dx(sinx/cosx)# =#1/cos^2x# #(cosx*cosx-sinx*(-sinx))# =#(cos^2x+sin^2x)/cos^2x# =#1/cos^2x#=#sec^2x#

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Answer 2

Emily Ammerman

The derivative of ( \frac{12}{\sin(x)} + \frac{1}{\cot(x)} ) is ( -\frac{12\cos(x)}{\sin^2(x)} - \csc^2(x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.