Implicit Differentiation is an application of the Chain Rule:

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

(31a) Determine dy/dx for 6x² - 3y² = 5

d(6x²)/dx - d(3y²)/dx = d(5)/dx

d(6x²)/dx - d(3y²)/dy • dy/dx = d(5)/dx
... differentiating implicitly ... Chain Rule

12x - 6y • dy/dx = 0

- 6y • dy/dx = -12x

dy/dx = 2x/y

(31b) Determine dy/dx for y³ + x²y - 2x² = 0

d(y³)/dx + d(x²y)/dx - d(2x²)/dx = d(0)/dx

d(y³)/dx + y • d(x²)/dx + x² • d(y)/dx - d(2x²)/dx = d(0)/dx
... d(uv)/dx = u • d(v)/dx + v • d(u)/dx ... Product Rule

d(y³)/dy • dy/dx + y • d(x²)/dx + x² • d(y)/dy • dy/dx - d(2x²)/dx = d(0)/dx
... differentiating implicitly ... Chain Rule

3y² • dy/dx + 2xy + x² • dy/dx - 4x = 0

(3y² + x²) • dy/dx + 2xy - 4x = 0 ... collecting like terms

dy/dx = (4x - 2xy)/(3y² + x²)

(32) Determine the slope of the curve 2x³ + 2y³ - 9xy = 0 at the point (1,2)

d(2x³)/dx + d(2y³)/dx - d(9xy)/dx< = d(0)/dx

d(2x³)/dx + d(2y³)/dx - (9x • d(y)/dx + y • d(9x)/dx) = d(0)/dx
... d(uv)/dx = u • d(v)/dx + v • d(u)/dx ... Product Rule

d(2x³)/dx + d(2y³)/dy • dy/dx - 9x • d(y)/dy • dy/dx - y • d(9x)/dx = d(0)/dx
... differentiating implicitly ... Chain Rule

6x² + 6y² • dy/dx - 9x • dy/dx - 9y = 0

(6y² - 9x) • dy/dx + 6x² - 9y = 0 ... collecting like terms

dy/dx = (9y - 6x²)/(6y² - 9x)

dy/dx = (3y - 2x²)/(2y² - 3x) ... reducing to lowest terms

At P(x,y) = (1,2) the slope equals:

dy/dx = [3(2) - 2(1)²] ÷ [2(2)² - 3(1)] = [6 - 2] ÷ [8 - 3] = 4/5

(33a-i) Given the relation 4x² + y² = 16 solve for y explicitly as a function of x, then differentiate with respect to x.

The relation may be written as two functions:

(33a-ii) Use Implicit Differentiation to differentiate 4x² + y² = 16 with respect to x

d(4x²)/dx + d(y²)/dx = d(16)/dx

d(4x²)/dx + d(y²)/dy • dy/dx = d(16)/dx
... differentiating implicitly ... Chain Rule

8x + 2y • dy/dx = 0

2y • dy/dx = -8x

dy/dx = - 4x/y

(34a) Find the derivative for the function y = (2x² - x)(3x² + 5)⁸

Let 3x² + 5 = u , therefore du/dx = 6x

Let 2x² - x = v , therefore dv/dx = 4x - 1

Let (3x² + 5)⁸ = w , where w = u⁸ , therefore dw/dx = d(u⁸)/du • du/dx
... invoking the Chain Rule and substituting for w

y = vw

dy/dx = d(vw)/dx = v • d(w)/dx + w • d(v)/dx ... Product Rule

dy/dx = d(vw)/dx = v • d(u⁸)/du • d(u)/dx + w • d(v)/dx
... invoking the Chain Rule and substituting for w

dy/dx = (2x² - x) • 8(3x² + 5)⁷ • 6x + (3x² + 5)⁸ • (4x - 1)
... substituting the original functions in terms of x

dy/dx = (3x² + 5)⁷ [48x(2x² - x) + (3x² + 5)(4x - 1)]
... factoring the common term (3x² + 5)⁷

dy/dx = (3x² + 5)⁷ [96x³ - 48x² + 12x³ + 20x - 3x² - 5]
... expanding the expression in the parenthesis

dy/dx = (3x² + 5)⁷ [108x³ - 51x² + 20x - 5]
... collecting like terms

(34b) Find the derivative for the function y = (t² - 6t) ÷ (t² + 6t)

y = [t(t - 6) ÷ t(t + 6)] = (t - 6) ÷ (t + 6)
... simplifying the expression and reducing it in degree

Let t - 6 = u , therefore du/dt = 1

Let 1/(t + 6) = (t + 6) ^-1 = v ... changing the division to a multiplication and permitting the use of the simpler Product Rule
Therefore dv/dt = (-1) • (t + 6) ^-2 = -1/(t + 6)²

Let (t - 6) ÷ (t + 6) = w = uv , therefore y = w⁵

dy/dt = d(w⁵)/dw • dw/dt = d(w⁵)/dw • d(uv)/dt ... Chain Rule
        = d(w⁵)/dw • [u • d(v)/dt + v • d(u)/dt] ... Product Rule

dy/dt = 5[(t - 6) ÷ (t + 6)]⁴ • [- (t - 6)/(t + 6)² + 1/(t + 6) • 1]
... substituting the original functions in terms of x

dy/dt = 5[(t - 6) ÷ (t + 6)]⁴ • [(- t + 6)/(t + 6)² + (t + 6)/(t + 6)²]
... multiplying the final term in the parenthesis by (t + 6)/(t + 6) , creating a common denominator for both terms

dy/dt = 5[(t - 6) ÷ (t + 6)]⁴ • [(- t + 6 + t + 6)/(t + 6)²]

dy/dt = 5[(t - 6)⁴ ÷ (t + 6)⁴] • [12 ÷ (t + 6)²]

dy/dt = 60(t - 6)⁴ ÷ (t + 6)⁶

Find the derivative for the function y = (3 - 2x)⁵(x² - 2)⁴

Let (3 - 2x)⁵ = u

du/dx = d(3 - 2x)⁵/d(3 - 2x) • d(3 - 2x)/dx = 5(3 - 2x)⁴ • (-2)
... Chain Rule

Let (x² - 2)⁴ = v

dv/dx = d(x² - 2)⁴/d(x² - 2) • d(x² - 2)/dx = 4(x² - 2)³ • (2x)
... Chain Rule

y = (uv)

dy/dx = d(uv)/dx = u • d(v)/dx + v • d(u)/dx ... Product Rule

dy/dx = (3 - 2x)⁵ • 4(x² - 2)³ • 2x + (x² - 2)⁴ • 5(3 - 2x)⁴ • (-2)
... substituting in the Product Rule formula

dy/dx = 2(3 - 2x)⁴ (x² - 2)³ • [4x(3 - 2x) - 5(x² - 2)]
... collecting terms and factoring

dy/dx = 2(3 - 2x)⁴ (x² - 2)³ (12x - 13x² + 10)
... solving the terms in the square parenthesis

Find the derivative for the function