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zeesbrat3

  • one year ago

Determine the slope of the graph of x4 = ln(xy) at the point (1, e).

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  1. freckles
    • one year ago
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    find the slope by first differentiating both sides of the given equation

  2. zeesbrat3
    • one year ago
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    But hot can you find the derivative of e?

  3. zeesbrat3
    • one year ago
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    @freckles

  4. freckles
    • one year ago
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    why do you want the derivative of e? by the way e is a constant and derivative of e is 0. But why do you need this information?

  5. freckles
    • one year ago
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    \[\frac{d}{dx}x^4=? \\ \frac{d}{dx}\ln(xy)=?\] the first can be found by using power rule second can be found using chain rule and product rule

  6. zeesbrat3
    • one year ago
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    Sorry, I'm just confused as to how to solve for the answer.. This question is on a practice test so I need to learn how to do it for the actual test.

  7. zeesbrat3
    • one year ago
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    So the first would be \[4x^3\] And the second would be \[\frac{ d }{ dx } \ln x + \ln y\]

  8. freckles
    • one year ago
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    oh okay so you used product rule for log but you still need to differentiate that sum: \[\frac{d}{dx}(\ln(x)+\ln(y)) \\ =\frac{d}{dx}\ln(x)+\frac{d}{dx}\ln(y)\] =?

  9. zeesbrat3
    • one year ago
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    \[\frac{ y }{ x }\]?

  10. freckles
    • one year ago
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    do you know that: where u is a function of x that: \[\frac{d}{dx}\ln(u)= \frac{u'}{u}\]

  11. freckles
    • one year ago
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    \[\frac{d}{dx}\ln(x)=\frac{(x)'}{x}=\frac{1}{x} \\ \frac{d}{dx}\ln(y)=\frac{y'}{y}\]

  12. freckles
    • one year ago
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    \[x^4=\ln(xy) \\ \text{ after differentiating both sides we have } \\ 4x^3=\frac{1}{x}+\frac{y'}{y} \] now you can replace x with 1 and y with e and solve for y' (the slope at (1,e))

  13. zeesbrat3
    • one year ago
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    So \[4x^3 = 1 + \frac{ y' }{ e }\]

  14. freckles
    • one year ago
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    close you need to replace all the x's with 1

  15. zeesbrat3
    • one year ago
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    Ooo whoops! \[4 = 1 + \frac{ y' }{ }\]

  16. zeesbrat3
    • one year ago
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    theres supposed to be an e under the y prime

  17. freckles
    • one year ago
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    \[4=1+\frac{y'}{e}\] yep now just solve for y' (this will be the slope at (1,e) since we replaced x with 1 and y with e)

  18. zeesbrat3
    • one year ago
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    So \[3e = y' ?\]

  19. freckles
    • one year ago
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    yep

  20. zeesbrat3
    • one year ago
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    Thank you! Would you mind helping me with one more? Logs through me for a loop

  21. freckles
    • one year ago
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    sure

  22. zeesbrat3
    • one year ago
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    Thanks! \[\frac{ d }{ dx }[x^3e^x] = x^2e^x(3x+2)\]

  23. zeesbrat3
    • one year ago
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    I have to find whether that is true or not

  24. freckles
    • one year ago
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    well do you know product rule?

  25. freckles
    • one year ago
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    x^3e^x is a product of x^3 and e^x

  26. freckles
    • one year ago
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    \[\frac{d}{dx}(x^3e^x)=e^x \frac{d}{dx}x^3+x^3 \frac{d}{dx}e^x\] what are the following: \[\frac{d}{dx}x^3=? \\ \frac{d}{dx}e^x=?\]

  27. zeesbrat3
    • one year ago
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    Right. So the first is \[3x^2\] and the second is 0?

  28. freckles
    • one year ago
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    the second is not 0 e^x is not a constant

  29. freckles
    • one year ago
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    for example e^2 and e^3 aren't the same so there is no way e^x remains the same

  30. freckles
    • one year ago
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    the derivative of e^x is e^x

  31. zeesbrat3
    • one year ago
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    Because it is a constant?

  32. freckles
    • one year ago
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    I thought I just said it wasn't a constant and gave a reason why it wasn't a constant. e is a constant e^x is not a constant (it varies) (Example: e^2 is not the same as e^3)

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