`arctan(xy) = arcsin(x + y) , (0,0)` Use implicit differentiation to find an equation of the tangent line at the given point

`arctan(xy) =arcsin(x + y)`

First, take the derivative of both sides of the equation using implicit differentiation.

`d/(dx)[arctan(xy)] = d/dx[arcsin(x + y)]`

Take note that the derivative formula of arctangent is

• 
  `d/dx[arctan(u)]=1/(1+u^2)*(du)/dx`

And the derivative formula of arcsine is

• 
  `d/dx[arcsin(u)] = 1/sqrt(1-u^2)*(du)/dx`

Applying these two formulas, the equation becomes

`1/(1+(xy)^2)*d/dx(xy) = 1/sqrt(1 - (x+y)^2)*d/dx(x+y)`

To take the derivative of xy, apply the product rule.

• 
  `d/dx (u * v) = u *(dv)/dx + v *(du)/dx`

Applying this formula, the equation becomes

`1/(1+(xy)^2)*(x*d/dx (y) + y*d/dx(x))= 1/sqrt(1 - (x+y)^2)*(d/dx(x)+d/dx(y))`

`1/(1+(xy)^2)*(x*(dy)/dx + y*1)= 1/sqrt(1 - (x+y)^2)*(1+(dy)/dx)`

Then, isolate `(dy)/dx` .

`x/(1+(xy)^2)*(dy)/dx +y/(1+(xy)^2)=1/sqrt(1 - (x+y)^2) +1/sqrt(1 - (x+y)^2)*(dy)/dx`

`x/(1+(xy)^2) *(dy)/dx - 1/sqrt(1-(x+y)^2)*(dy)/dx = 1/sqrt(1-(x+y)^2) - y/(1+(xy)^2)`

`(x/(x+(xy)^2)-1/sqrt(1-(x+y)^2))*(dy)/dx= 1/sqrt(1-(x+y)^2) - y/(1+(xy)^2)`

`(dy)/dx =(1/sqrt(1-(x+y)^2) - y/(1+(xy)^2))/(x/(1+(xy)^2) - 1/sqrt(1-(x+y)^2))`

`(dy)/dx =(1/sqrt(1-(x+y)^2) - y/(1+(xy)^2))/(x/(1+(xy)^2) - 1/sqrt(1-(x+y)^2)) * (((1+(xy)^2)sqrt(1-(x+y)^2))/1)/(((1+(xy)^2)sqrt(1-(x+y)^2))/1)`

`(dy)/dx = ((1+(xy)^2)-ysqrt(1-(x+y)^2))/(xsqrt(1-(x+y)^2)-(1+(xy)^2))`

`(dy)/dx = (1+(xy)^2 - ysqrt(1-(x+y)^2))/(xsqrt(1-(x+y)^2 )-1-(xy)^2)`

Then, plug-in the given point to get the slope of the curve at that point. The given point is (0,0).

`(dy)/dx= (1+(0*0)^2 -0*sqrt(1 - (x+y)^2))/(0*sqrt(1-(0+0)^2)-1-(0*0)^2)=(1+0+0)/(0-1-0)=1/(-1)=-1`

Take note that the slope of a curve at point (x,y) is equal to the slope of the line tangent to that point.  So the slope of the tangent line is

`m=(dy)/dx = -1`

Then, apply the point-slope form to get the equation of the line.

`y - y_1 = m(x - x_1)`

Plugging in the values, it becomes

`y - 0=-1(x - 0)`

`y = -1(x)`

`y=-x`

Therefore, the equation of the tangent line is `y = -x` .