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Jaweria

  • 2 years ago

Find dy/dx: 4x^2+3xy^2-6x^2y=y^3.

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  1. abb0t
    • 2 years ago
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    Using implicit differentiation with respect to x. This means that you take the derivative of everything, even y, but everytime you take the deriviative of y, you multiply it by y'. For instance:\[\frac{ dy }{ dx }(x+y) = 1+\frac{ dy }{ dx } = 1+y'\] then, using your algebra skills, rearrange to find dy/dx

  2. Jaweria
    • 2 years ago
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    ahan ok

  3. abb0t
    • 2 years ago
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    Remember: \[3xy^2 = 3x \frac{ d }{ dx}y^2+y^2\frac{ d }{ dx }3x\]

  4. Jaweria
    • 2 years ago
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    I m little bit confuse here

  5. cwrw238
    • 2 years ago
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    what you are doing is treating y as a function of x (which is implied in the expression)

  6. abb0t
    • 2 years ago
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    Of course!

  7. Jaweria
    • 2 years ago
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    thanks :-)

  8. abb0t
    • 2 years ago
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    Start by taking the derivative as you normally would for the left side of the function. And like @cwrw238 pointed out, you're treating y as a function of x.

  9. Jaweria
    • 2 years ago
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    ok

  10. abb0t
    • 2 years ago
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    If it helps, you can break it down to see it more clearly: \[\frac{ d }{ dx }4x^2 =\] \[\frac{ d }{ dx }3xy^2 = (3x \times \frac{ d }{ dx }y^2)+(y^2\frac{ d }{ dx }3x) = [3x \times 2y \frac{ dy }{ dx }]+[y^2 \times 3]\] \[\frac{ d }{ dx }6x^2y = \] NOW, the right side: \[\frac{ d }{ dx }y^3 = 3y^2\frac{ dy }{ dx } \]

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