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math_proof

  • 3 years ago

find partial derivative Hx and Hy of H(x,y)=(y^2+1)e^x

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  1. cruffo
    • 3 years ago
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    can you explain how you find the partial derivative with respect to x? Short sentance...

  2. math_proof
    • 3 years ago
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    you basically do derivative just with respect to x and treating y as a constant

  3. zepdrix
    • 3 years ago
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    So Mproof, Hmm If you take partials, you'll be treating the OTHER variable as a constant, meaning you won't have the product rule as it might seem at first glance. Does that help? :O

  4. math_proof
    • 3 years ago
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    like 3x^2y+2 fx=6xy and Fy is 3x^2

  5. cruffo
    • 3 years ago
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    right. So what is confusing you about this problem?

  6. math_proof
    • 3 years ago
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    would Hx be 0?

  7. cruffo
    • 3 years ago
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    no. What is the regular derivative of f(x) = e^x?

  8. math_proof
    • 3 years ago
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    same

  9. zepdrix
    • 3 years ago
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    Think of the equation as Ce^x when taking the partial WRT x. Maybe that will help :)

  10. math_proof
    • 3 years ago
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    but don't you have to take a derivative of the (y^2+1) with respect to x?

  11. zepdrix
    • 3 years ago
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    no, thats just a constant attached to e^x :d

  12. zepdrix
    • 3 years ago
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    Maybe one thing you can do to convince yourself is, distribute the e^x to each term in the brackets. Then think about what you have :o

  13. math_proof
    • 3 years ago
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    so the derivative with respect to x will be the same as the given problem

  14. zepdrix
    • 3 years ago
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    ya :) good

  15. math_proof
    • 3 years ago
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    ooo I get it

  16. math_proof
    • 3 years ago
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    with respect to Y would it be 2ye^x?

  17. zepdrix
    • 3 years ago
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    (y^2 + 1)e^x = y^2 e^x + e^x Hy = 2y e^x + 0 Yes, very good ^^

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