What is the derivative of this function #y= (1+sin x)/(1-sin x)#?

1 Answer
Steve M
Nov 5, 2016

# dy/dx = ( 2cosx ) / (1-sinx)^2#

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #, or less formally, # (u/v)' = (v(du)-u(dv))/v^2 #

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with # y=(1+sinx)/(1-sinx) # Then

# { ("Let "u=1+sinx, => , (du)/dx=cosx), ("And "v=1-sinx, =>, (dv)/dx=-cosx ) :}#

# :. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
# :. dy/dx = ( (1-sinx)(cosx) - (1+sinx)(-cosx) ) / (1-sinx)^2#
# :. dy/dx = ( cosx-sinxcosx + cosx +sinxcosx ) / (1-sinx)^2#
# :. dy/dx = ( 2cosx ) / (1-sinx)^2#

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