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

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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 $(pi/3,pi/3)$ :

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

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

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

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

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