Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 79

arcsin(x) +arcsin(y) = pi/2
First, take the derivative of both sides of the equation using implicit differentiation.
d/dx[arcsin(x) + arcsin(y)] = d/dx (pi/2)
d/dx[arcsin(x)] + d/dx[arcsin(y)]=d/dx(pi/2)
Take note that the derivative formula of arcsine is
d/dx[arcsin(u)] = 1/sqrt(1-u^2)*(du)/dx
And the derivative of a constant is zero.
d/dx(c) = 0
Applying these two formulas, the equation becomes
1/sqrt(1-x^2) *d/dx(x) + 1/sqrt(1-y^2)*d/dx(y) = 0
1/sqrt(1-x^2) *1 + 1/sqrt(1-y^2)*(dy)/dx=0
1/sqrt(1-x^2) + 1/sqrt(1-y^2)*(dy)/dx = 0
Then, isolate (dy)/dx .
1/sqrt(1-y^2)*(dy)/dx = -1/sqrt(1-x^2)
(dy)/dx = -1/sqrt(1-x^2)*sqrt(1-y^2)/1
(dy)/dx = -sqrt(1-y^2)/sqrt(1-x^2)
Then, plug-in the given point to get the slope of the curve on that point. The given point is (sqrt2/2,sqrt2/2) .
(dy)/dx = -sqrt(1- (sqrt2/2)^2)/sqrt(1-(sqrt2/2)^2)=-1
Take note that the slope of a curve at point (x,y) is the slope of the line tangent to that point. Hence, the slope of the tangent line is
m = (dy)/(dx) = -1
Now that the slope of line that is tangent to the graph of function at (sqrt2/2,sqrt2/2) is known, 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 - sqrt2/2=-1(x - sqrt2/2)
y-sqrt2/2=-x + sqrt2/2
y = -x+sqrt2/2+sqrt2/2
y = -x + (2sqrt2)/2
y = -x + sqrt2

Therefore, the equation of the tangent line is y = -x + sqrt2 .