Find the equation of the tangent line to the given equation at the indicated point
a) y^2=cos(5x-3y) at ((2pi+3)/5, 1)
b) x^2+x^2y+4=0 at (2,-2)
For this type of problem , the first thing you should do is to make sure that the given point actually lies on the curve, or else whatever we do would be just gibberish.
I did this, and both points satisfy their matching equation.
So, we take the derivative, plug in the point to get the slope.
Now you have the slope and a point, so using your grade 9 method of finding the equation, just find the equation of the tangent.
y^2 = cos(5x - 3y)
2y dy/dx = -sin(5x-3y)*(5 - 3dy/dx)
plug in the given point:
2 dy/dx = -sin(5(2π+3)/5 - 3)*(5(2π+3)/5 - 3dy/dx)
2dy/dx = -sin(2π)*(2π+3 - 3dy/dx)
2dy/dx = -(0)(2π + 3 - 3dy/dx)
2 dy/dx = 0
dy/dx = 0 <---- a horizontal line
equation:
y = 1
check my arithmetic, easy to make errors if you don't write it down first
2nd:
x^2+x^2y+4=0 at (2,-2)
2x + x^2(dy/dx) + 2xy = 0
for the given point:
4 + 4dy/dx - 8 = 0
4dy/dx = 4
dy/dx = 1 <--- the slope
equation of tangent at (2,-2):
y+2 = 1(x-2)
y = x - 4