using implicit differentiation find the equation of the tangent line to the graph of the following function at the indicated point x^2 y^3 -y^2+xy-1=0 at (1,1)
x^2(3y^2dy) + 2x dx(y^3) - 2ydy +xdy + dx y = 0
dy (3x^2y^2-2y +x) +dx(2xy^3+y)=0
dy (3-2+1) + dx (2+1) = 0
2 dy = -3dx
m = slope = dy/dx = -3/2
so y = -3/2 x + b
1 = -3/2 + b
b = 5/2
so y = -3/2 x + 5/2
or
2 y = -3 x + 5
To find the equation of the tangent line to the graph of the function at a given point using implicit differentiation, follow these steps:
Step 1: Differentiate both sides of the equation with respect to x, treating y as a function of x.
The given equation is:
x^2 y^3 - y^2 + xy - 1 = 0
Differentiating both sides with respect to x:
d/dx(x^2 y^3 - y^2 + xy - 1) = d/dx(0)
2xy^3 + x^2 * 3y^2(dy/dx) - 2y(dy/dx) + x + y = 0
Step 2: Substitute the coordinates of the given point (1, 1) into the differentiated equation.
Substituting x = 1 and y = 1 into the equation:
2(1)(1^3) + (1^2)(3)(1^2)(dy/dx) - 2(1)(dy/dx) + 1 + 1 = 0
2 + 3(dy/dx) - 2(dy/dx) + 2 = 0
Simplifying the equation:
3(dy/dx) - 2(dy/dx) = -4
(dy/dx) = -4
Step 3: Using the derivative dy/dx, find the slope of the tangent line at the given point.
The slope of the tangent line at the point (1, 1) is -4.
Step 4: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Using the point (1, 1) and the slope -4:
y - 1 = -4(x - 1)
Expanding and rearranging the equation:
y - 1 = -4x + 4
y = -4x + 5
Therefore, the equation of the tangent line to the graph of the function at the point (1, 1) is y = -4x + 5.