What is the slope of the tangent line of $tan(2x)/tan^2(xy)= C$ , where C is an arbitrary constant, at $(pi/3,pi/3)$ ?

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Answer 1 · 2017-07-05T17:57:09

$dy/dx=-1.89585$

At the point $(pi/3,pi/3)$ , we get

$tan(2pi/3)/tan^2(pi^2/9)=C$

$=> -sqrt(3)/tan^2(pi^2/9)=C$

$=> C~~-0.45623$

So then we can rewrite $C$ on the right hand side like this

$tan(2x)/tan^2(xy)=-0.45623$

Cross multiply

$tan(2x)=-0.45623tan^2(xy)$

Next, we can implicitly differentiate each side

$d/dx(tan(2x))=-0.45623d/dx(tan^2(xy))$

$2sec^2(2x)=-0.45623(2tan(xy)sec^2(xy)(x(dy)/(dx)+y))$

Now we can plug in the point $(pi/3,pi/3)$ and solve for $dy/dx$

$2sec^2(2pi/3)=-0.45623(2tan(pi/3*pi/3)sec^2(pi/3*pi/3)(pi/3(dy)/(dx)+pi/3))$

After a little work with a calculator, you get

$dy/dx=-1.89585$

What is the slope of the tangent line of tan(2x)/tan^2(xy)= C tan(2x)/tan^2(xy)= C , where C is an arbitrary constant, at (pi/3,pi/3)(pi/3,pi/3)?