36.1k views
0 votes
Differentiate the following Functions
5x^2-2xy + 4y^3= 5

User Radolino
by
7.4k points

1 Answer

4 votes

Answer:


\displaystyle y' = (y - 5x)/(x + 6y^2)

General Formulas and Concepts:

Algebra I

  • Terms/Coefficients
  • Factoring

Calculus

Differentiation

  • Derivatives
  • Derivative Notation
  • Implicit Differentiation

Derivative Property [Multiplied Constant]:
\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:
\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
\displaystyle (d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Derivative Rule [Chain Rule]:
\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)

Explanation:

Step 1: Define

Identify


\displaystyle 5x^2 - 2xy + 4y^3 = 5

Step 2: Differentiate

  1. Implicit Differentiation:
    \displaystyle (dy)/(dx)[5x^2 - 2xy + 4y^3] = (dy)/(dx)[5]
  2. Rewrite [Derivative Property - Addition/Subtraction]:
    \displaystyle (dy)/(dx)[5x^2] - (dy)/(dx)[2xy] + (dy)/(dx)[4y^3] = (dy)/(dx)[5]
  3. Rewrite [Derivative Property - Multiplied Constant]:
    \displaystyle 5(dy)/(dx)[x^2] - 2(dy)/(dx)[xy] + 4(dy)/(dx)[y^3] = (dy)/(dx)[5]
  4. Basic Power Rule [Chain Rule]:
    \displaystyle 10x - 2(dy)/(dx)[xy] + 12y^2y' = 0
  5. Product Rule:
    \displaystyle 10x - 2\bigg[ (dy)/(dx)[x]y + x(dy)/(dx)[y] \bigg] + 12y^2y' = 0
  6. Basic Power Rule [Chain Rule]:
    \displaystyle 10x - 2\bigg[ y + xy' \bigg] + 12y^2y' = 0
  7. Simplify:
    \displaystyle 10x - 2y + 2xy' + 12y^2y' = 0
  8. Isolate y' terms:
    \displaystyle 2xy' + 12y^2y' = 2y - 10x
  9. Factor:
    \displaystyle y'(2x + 12y^2) = 2y - 10x
  10. Isolate y':
    \displaystyle y' = (2y - 10x)/(2x + 12y^2)
  11. Factor:
    \displaystyle y' = (2(y - 5x))/(2(x + 6y^2))
  12. Simplify:
    \displaystyle y' = (y - 5x)/(x + 6y^2)

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

User Ikhvjs
by
7.1k points