What is the slope of the tangent line of #xarctan(y/pi)= C #, where C is an arbitrary constant, at #(pi/3,pi/3)#?

Question

1261 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-19T02:54:12

#dy/dx=-10/9arctan(1//3)#

Taking the derivative, we first use the product rule:

#(d/dxx)arctan(y/pi)+x(d/dxarctan(y/pi))=0#

The typical arctangent derivative is #d/dxarctan(x)=1/(1+x^2)#, so here we apply the chain rule:

#arctan(y/pi)+x(1/(1+(y/pi)^2))(d/dxy/pi)=0#

The derivative with respect to #x# of #y/pi#, which is the linear equation #1/piy#, so its derivative is #1/pidy/dx#:

#arctan(y/pi)+x/(1+y^2/pi^2)(1/y)dy/dx=0#

Solving for #dy/dx# at #(pi/3,pi/3)#:

#arctan(1//3)+(pi//3)/(1+(pi^2//9)//pi^2)(1/(pi//3))dy/dx=0#

#arctan(1//3)+(1/(1+1//9))dy/dx=0#

#arctan(1//3)+9/10(dy/dx)=0#

#dy/dx=-10/9arctan(1//3)#