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

2 Answers
Jul 16, 2018

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

Explanation:

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#

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

Explanation:

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#