What is the slope of the tangent line of this equation?

Question

the equation is cos(x+1) + 1 = sin(xy²)
the x-coordinate is 2 and the y-coordinate is 1.8.

1) find the slope of the tangent line.
2) give an equation of the tangent line and solve it for y.

the x coordinate 2 and y coordinate 1.8 is the point of tangency of the graph.

Answer 1

implicit derivate
-sin(x+1)= cos(xy^2) *( y^2+2xy*y´)
Accepting the given values at x= 2 and y m=1.8 so
-sin3 = cos(2*3.24) *( 3.24+7.2*y´)so y´= -0.4696
y-1.8= -0.4696(x-2) so
y=-0.4696x+2.7392
But if you take the given equation and put in given x and y
cos(3)+1 = -0.99+1=0.01
sin(2*1.8^2) =0.1955 so the given coordinates don´t satisfy the equation
If x= 2
0.01 = sin(2y^2) so2y^2 = 0.01 ( for small angles the arc and the sin are almost equal)
so y^2=0.005 and y =+-0.0707
Taking one point (2,0.0707)
-sin3 = 0.9999*(0.005+0.1414 y´) so y´= 0.1058 which is the
real slope at x=2
Now you can do what I did before with y´=-0.4696
Check your problem

Answer 2

The point (2,1.8) doesn't even belong to the curve. Did you mistype somewhere?

Answer 3

http://en.wikipedia.org/wiki/Tangent_line#Calculus