`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` .
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