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anonymous

  • one year ago

Use implicit differentiation to find the slope of the tangent line to the curve2xy^3+5xy=35 at the point (5,1).

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  1. zepdrix
    • one year ago
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    Hey burns :) What are you having trouble with? We need to find y' at the specified point, ya?

  2. zepdrix
    • one year ago
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    \[\Large\rm 2xy^3+5xy=35 \]Differentiating:\[\Large\rm \color{royalblue}{(2x)'}y^3+2x\color{royalblue}{(y^3)'}+\color{royalblue}{(5x)'}y+5x\color{royalblue}{(y)'}=\color{royalblue}{(35)'}\]We would have to apply the product rule in both cases, do you understand this setup?

  3. anonymous
    • one year ago
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    How did you get that set up? i know the product rule, I am just unfimiliar with how to get it set up properly

  4. zepdrix
    • one year ago
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    To get more comfortable with product rule, I would recommend doing the "setup" as I did above, eventually you can skip that step when you feel good at these. Example:\[\Large\rm (xy)'=x'y+xy'\] If differentiating with respect to x, the x' becomes 1 simply,\[\Large\rm (xy')=1\cdot y+xy'\]

  5. anonymous
    • one year ago
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    For implicit differentiation, just treat x and y as regular variables for derivatives. For every y that you derive, you need to add a y' after derivation,

  6. zepdrix
    • one year ago
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    \[\Large\rm (\color{orangered}{2x}\color{royalblue}{y^3})'=(\color{orangered}{2x})'\color{royalblue}{y^3}+\color{orangered}{2x}(\color{royalblue}{y^3})'\]Ok with this setup for the first term? :o

  7. zepdrix
    • one year ago
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    You can group the 2 with either the x OR the y^3, it doesn't matter which, just be consistent.

  8. zepdrix
    • one year ago
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    So after we have our setup, we apply our differentiation,\[\Large\rm (\color{orangered}{2x}\color{royalblue}{y^3})'=(2)\color{royalblue}{y^3}+\color{orangered}{2x}(3y^2y')\]

  9. zepdrix
    • one year ago
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    Having any confusion as to why we have y' showing up, but not x'? :O

  10. anonymous
    • one year ago
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    Is it because we canceled out the x's?

  11. zepdrix
    • one year ago
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    Ummmm, well it's because of the stuff that we end up with:\[\Large\rm \frac{d}{dx}x^2=2x\frac{d}{dx}x\]Here is a simple chain rule being applied ^ We differentiated the outermost function, and then we multiply by the derivative of the inner function,\[\Large\rm \frac{d}{dx}x^2=2x\frac{dx}{dx}\]But this thing really has no significance, that's why they usually ignore doing this chain when you're learning your power rule and such. It's like.. if you were to read into it, the numerator says "how much does x change" with the denominator "as x changes an immeasurably small amount?"\[\Large\rm \frac{d}{dx}x^2=2x\cdot 1\]

  12. zepdrix
    • one year ago
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    I dunno, maybe that's an unnecessary and long-winded explanation, i'm just trying to explain that we're always apply the chain rule in this way, but it only actually matters when it's a variable that doesn't match our operator variable.

  13. zepdrix
    • one year ago
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    So by comparison,\[\Large\rm \frac{d}{dx}y^2=2y\frac{d}{dx}y\]giving us,\[\Large\rm \frac{d}{dx}y^2=2y\frac{dy}{dx}\]or\[\Large\rm \frac{d}{dx}y^2=2y y'\]

  14. zepdrix
    • one year ago
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    So just to review that first term again, setup,\[\Large\rm (2xy^3)'=(2x)'y^3+2x(y^3)'\]differentiate,\[\Large\rm (2xy^3)'=(2x')y^3+2x(3y^2y')\]and simplify,\[\Large\rm (2xy^3)'=(2)y^3+2x(3y^2y')\]

  15. anonymous
    • one year ago
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    and then plug in the points?

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