How do you use implicit differentiation to find dy/dx given #sinx+cosy=0#?

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Answer 1 · 2016-12-21T19:19:49

#(dy)/(dx) = 1#

First you differentiate the expression, considering that based on the chain rule:

#(dcosy)/(dx) = (dcosy)/(dy)*(dy)/(dx) #

so that you get:

#d/(dx)(sinx+cosy) = cosx-siny(dy)/(dx) = 0#

and resolving for #(dy)/(dx) #

#(dy)/(dx) = cosx/siny#

or

#(dy)/(dx) = cosx/sqrt(1-cos^2y)#

Now we can obtain #cosy# from the original equation:

#cosy = -sinx#

and then:

#(dy)/(dx) = cosx/sqrt(1-sin^2x)= cosx/cosx = 1#

In fact, we can have:

#cosy = -sinx#

for any value of #x# only if #y=x+pi/2+2kpi#