How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $sin x = x (1 + tan y)$ $sinx=x(1+tany)$ ?

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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:

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

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