Implicit Differentiation is an application of the Chain Rule:

d(kyⁿ)/dx = d(kyⁿ)/dy · dy/dx = nky^n-1 · dy/dx

(35a) Determine dy/dx for 9x² - 16y² = 1

d(9x²)/dx - d(16y²)/dx = d(1)/dx

d(9x²)/dx - d(16y²)/dy • dy/dx = d(1)/dx
... differentiating implicitly ... Chain Rule

18x - 32y • dy/dx = 0

-32y • dy/dx = -18x

dy/dx = - 9x/16y

(35b) Determine dy/dx for y³ + 5xy + x³ = 1

d(y³)/dx + d(5xy)/dx + d(x³)/dx = d(1)/dx

d(y³)/dx + y • d(5x)/dx + 5x • d(y)/dx + d(x³)/dx = d(1)/dx
... d(uv)/dx = u • d(v)/dx + v • d(u)/dx ... Product Rule

d(y³)/dy • dy/dx + y • d(5x)/dx + 5x • d(y)/dy • dy/dx + d(x³)/dx = d(1)/dx
... differentiating implicitly ... Chain Rule

3y² • dy/dx + 5y + 5x • dy/dx + 3x² = 0

(3y² + 5x) • dy/dx = -5y - 3x² ... collecting like terms

dy/dx = -(5y + 3x²)/(3y² + 5x)

(36) Determine the slope of the curve 8x³ + 3xy + 8y³ = 19 at the point (1, 1)

d(8x³)/dx + d(3xy)/dx + d(8y³)/dx = d(19)/dx

d(8x³)/dx + y • d(3x)/dx + 3x • d(y)/dx + d(8y³)/dx = d(19)/dx
... d(uv)/dx = u • d(v)/dx + v • d(u)/dx ... Product Rule

d(8x³)/dx + y • d(3x)/dx + 3x • d(y)/dy • dy/dx + d(8y³)/dy • dy/dx = d(19)/dx
... differentiating implicitly ... Chain Rule

24x² + 3y + 3x • dy/dx + 24y² • dy/dx = 0

(3x + 24y²) • dy/dx = -3y - 24x² ... collecting like terms

dy/dx = - (y + 8x²)/(8y² + x) ... cancelling like terms

At P(x, y) = (1, 1) , slope = dy/dx = - (1 + 8(1)²)/(8(1)² + 1) = -9/9 = -1

(37) Solve the equation of the tangent to the curve x² - y² - x = 1 at (2, 1)

d(x²)/dx - d(y²)/dx - d(x)/dx = d(1)/dx

d(x²)/dx - d(y²)/dy • dy/dx - d(x)/dx = d(1)/dx
... differentiating implicitly ... Chain Rule

2x - 2y • dy/dx - 1 = 0

-2y • dy/dx = 1 - 2x

dy/dx = (1 - 2x)/(-2y) ... cancelling like terms

At P(x, y) = (2, 1) , slope = dy/dx = (1 - 2(2))/(-2(1)) = 3/2

(y - y₁) = m(x - x₁) ... point-slope form of the equation of the tangent

(y - 1) = 3/2(x - 2) ... substituting

y = 3x/2 - 2 ... equation of the tangent

The same solution is obtained by re-arranging the equation as y = (x² - x - 1)^1/2

dy/dx = d(x² - x - 1)^1/2/d(x² - x - 1) • d(x² - x - 1)/dx ... Chain Rule

dy/dx = (2x - 1) ÷ [2(x² - x - 1)^1/2]

Slope = dy/dx = (2(2) - 1) ÷ [2((2)² - 2 - 1)^1/2] = 3/2 ... substituting

(38) Determine the equation(s) of the tangent(s) to the curve xy + y = 8
having a slope of -1/2

Re-arrange the equation: y = 8/(x + 1)
dy/dx = d(8/(x + 1))/dx = -8/(x + 1)²

or, differentiating xy + y = 8 implicitly:
x • d(y)/dx • dy/dx + y • d(x)/dx + d(y)/dx • dy/dx = d(8)/dx
dy/dx = -y/(x + 1)
Since y = 8/(x + 1) , substituting for y: dy/dx = -8/(x + 1)²

The slope -1/2 is given, and the derivative also represents the slope.
Therefore -8/(x + 1)² = -1/2

Re-arranging the terms and expanding (x + 1)² :
x² - 2x - 15 = 0

The General Quadratic Formula returns x = 3 or x = -5

Sustituting these values in the original equation y = 8/(x + 1) returns the points:
P₁(x₁, y₁) = (3, 2) and P₂(x₂, y₂) = (-5, -2)

The points on the lines are known, and the slope of -1/2 is given.
The Point-Slope Equation for a straight line:
(y - y₁) = m(x - x₁)
(y - y₂) = m(x - x₂)

Substituting for x and y in the above formulas:
(y - 2) = -1/2(x - 3) , and y = -x/2 + 7/2
(y - (-2)) = -1/2(x - (-5)) , and y = -x/2 - 9/2

(39a) The second derivative of a function may be determined by taking the derivative of the derivative: dy(dy/dx)/dx = d²y/dx²

Example: If y = x³ + 2x² - x -2

dy/dx = 3x² + 4x -1

d²y/dx² = 6x + 4

If the first derivative or slope represents the rate of change for the function, the second derivative represents the rate of change of the rate of change, that is, how the slope of the function is changing.