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anlnt4

  • 3 years ago

Find dy/dx for y=ln(5−x)^6

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  1. waterineyes
    • 3 years ago
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    \[y = \ln(5-x)^6\] take derivative : \[\frac{dy}{dx} = 6 \cdot \ln(5-x)^5 \frac{d(5-x)}{dx}\]

  2. anlnt4
    • 3 years ago
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  3. waterineyes
    • 3 years ago
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    It is by chain rule.. In chain rule we are using power rule also which says : \[y = f(x)^n\] \[\frac{dy}{dx} = n \cdot f(x)^{n-1} \cdot \frac{d(f(x))}{dx}\]

  4. anlnt4
    • 3 years ago
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    Wait, isn't that in the case that the 6th power would be to the entire thing? I thought you brought the 6 in front since lna^b=blna

  5. waterineyes
    • 3 years ago
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    \[\frac{dy}{dx} = 6 \cdot \ln(5-x)^5 \frac{d(\ln(5-x)}{dx} \cdot \frac{d(5-x)}{dx}\]

  6. Algebraic!
    • 3 years ago
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    ln((x-5)^6) = 6ln(x-5) ( 6ln(x-5) ) ' = 6/(x-5)

  7. waterineyes
    • 3 years ago
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    Mixing up things...

  8. anlnt4
    • 3 years ago
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    oh! duh! Got it, thank you!!1

  9. waterineyes
    • 3 years ago
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    \[y = \ln(5-x)^6\] \[\frac{dy}{dx} = \frac{1}{(5-x)^6} \times 6 \times (5-x)^5 \times (-1)\]

  10. waterineyes
    • 3 years ago
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    Is this right???

  11. waterineyes
    • 3 years ago
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    It will reduce to : \[\frac{dy}{dx} = \frac{-1}{(5-x)}\]

  12. Algebraic!
    • 3 years ago
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    ln((5-x)^6) = 6ln(5-x) ( 6ln(5-x) ) ' = -6/(5-x)

  13. waterineyes
    • 3 years ago
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    Forgot 6 there.. Oh God...

  14. waterineyes
    • 3 years ago
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    \[\frac{dy}{dx} = \frac{-6}{(5-x)}\]

  15. waterineyes
    • 3 years ago
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    Yes by power rule, you can bring 6 in front: \[y = 6 \cdot \ln(5-x)\] Now take the derivative..

  16. anlnt4
    • 3 years ago
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    Okay, I've got it figured out! Thank you! If you'd like to help some more I just posted another question :)

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