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Study23 Group Title

Implicit Differentiation? Don't Understand it :( PLEASE HELP! Here's the problem I am looking at: \(\ \large x^3+x^2y+4x^2=6 .\) Find \(\ \frac{dy}{dx}. \) Please show me STEP by STEP!!

  • 2 years ago
  • 2 years ago

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  1. Ventricate Group Title
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    3x^2+2xy+2x dy/dx +8x= 0 then isolate dy/dx and simplify from there. Do you get that or need me to explain it?

    • 2 years ago
  2. Study23 Group Title
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    Could you explain it to me please?

    • 2 years ago
  3. Ventricate Group Title
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    Are you familiar with the derivative rules?

    • 2 years ago
  4. Study23 Group Title
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    Yes

    • 2 years ago
  5. Ventricate Group Title
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    Ok so the first part 3x^2 is from the normal derivative way and then when we move on to derive x^2y we have to use the product rule : (x^2)(y) The product rule I hope you know already is (a'(x)*b)+(a*b'(x)) for ab respectively. Do you understand up to here?

    • 2 years ago
  6. Study23 Group Title
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    How does the product rule look like when you derive \(\ x^2y?\)

    • 2 years ago
  7. Ventricate Group Title
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    It should be f'(x) *y + f(x) *y'. Derivative of x^2 is 2x and derivative of y is dy/dx.

    • 2 years ago
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    Could you show me the product rule in terms of \(\ x^2y \text{, please?}\)

    • 2 years ago
  9. Ventricate Group Title
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    Sure (2x)(y) +(x^2)(dy/dx)

    • 2 years ago
  10. Study23 Group Title
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    What does the dy/dx mean? dy/dx of x^2? or y?

    • 2 years ago
  11. Study23 Group Title
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    I'm REALLY Confused BY this! !

    • 2 years ago
  12. Ventricate Group Title
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    it's the derivative of y

    • 2 years ago
  13. Ventricate Group Title
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    Let's say you had y^3 you would derive it as \[3y^2 \frac{ dy }{ dx }\]

    • 2 years ago
  14. Study23 Group Title
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    Okay

    • 2 years ago
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