What is the derivative of #x + cos(x+y)=0#?

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Answer 1 · 2018-07-16T11:10:41

#y'=(1-sin(x+y))/sin(x+y)#

Assuming you mean #y=y(x)#
we get by the chain rule

#1-sin(x+y)-y'sin(x+y)=0#
so we get

#y'=(1-sin(x+y))/sin(x+y)# if #sin(x+y)ne 0#

Answer 2 · 2018-07-16T11:22:00

#dy/dx=\frac{1}{\sqrt{1-x^2}}-1#

Given equation:

#x+\cos(x+y)=0#

#cos(x+y)=-x#

#x+y=\pi-\cos^{-1}(x)#

#y=\pi-\cos^{-1}(x)-x#

Differentiating above equation w.r.t. #x# as follows

#\frac{dy}{dx}=\frac{d}{dx}(\pi-\cos^{-1}(x)-x)#

#=0-(-\frac{1}{\sqrt{1-x^2}})-1#

#=\frac{1}{\sqrt{1-x^2}}-1#