How do you find #dy/dx# by implicit differentiation given #lny=sinx#?

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Answer 1 · 2017-02-01T17:43:12

#(dy)/(dx) = ycosx#

Differentiate both sides with respect to #x#, keeping in mind that:

#d/(dx) f(y(x)) = f'(y(x))*y'(x)#

so:

#d/(dx) lny = d/(dx) sinx#

#1/y (dy)/(dx) = cosx#

#(dy)/(dx) = ycosx#

We can note that the original equation allows us also to express #y(x)# explicitly:

#lny = sinx <=> y=e^sinx#

so that:

#(dy)/(dx) = cosxe^sinx#