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#