What is the derivative of # d/dx (x+cosx ÷ tanx )#?
1 Answer
# d/dx (x+cosx ÷ tanx ) = 1 - cosx - cotxcscx #
Explanation:
# d/dx (x+cosx ÷ tanx ) = d/dx (x+cosx/tanx ) #
# " "= d/dx (x)+d/dx(cosx/tanx) #
# " "= 1 + { (tanx)(-sinx) - (cosx)(sec^2x) }/ (tanx)^2 #
# " "= 1 - (tanx)(sinx)/tan^2x - (cosx)(sec^2x)/(tanx)^2 #
# " "= 1 - (sinx)*cosx/sinx - (cosx)(1/cos^2x)*cos^2x/sin^2x #
# " "= 1 - cosx - (cosx)/sin^2x #
# " "= 1 - cosx - cotxcscx #
