How do you find #dy/dx# by implicit differentiation of #sinx=x(1+tany)#?

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Answer 1 · 2017-01-13T11:23:01

#y'(x) = (x(xcosx -sinx)) /(sin^2x-2xsinx+2x^2)#

Differentiate both sides of the equation with respect to #x#:

#d/(dx) (sinx) = d/(dx) (x(1+tany))#

remembering that:

#d/(dx) f(y(x)) = (df)/(dy)* (dy)/(dx)#

So:

#cosx = (1+tany) + (xy')/cos^2y#

Solving for #y'#:

#y'(x) = (cosx -1 -tany)cos^2y/x#

From the original equation:

# sinx = x(1+tany)#

we have that:

#tany = sinx/x-1#

and as:

#cos^2y = 1/sec^2y = 1/(1+tan^2y)#

#cos^2y = 1/(1+(sinx/x-1)^2) = x^2/(sin^2x-2xsinx+2x^2)#

We can make #y'(x)# explicit in #x# using these expressions:

#y'(x) = (cosx -1 -sinx/x+1)x^2/(sin^2x-2xsinx+2x^2) = (x(xcosx -sinx)) /(sin^2x-2xsinx+2x^2)#