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amyna

  • one year ago

diffrentiate: g(x)=4^3x^2 Thank you for your help!

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  1. amyna
    • one year ago
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    |dw:1441504840472:dw|

  2. triciaal
    • one year ago
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    differentiate power raised to a power think we can do substitution

  3. triciaal
    • one year ago
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    |dw:1441505133978:dw|

  4. Astrophysics
    • one year ago
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    Try logarithms

  5. triciaal
    • one year ago
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    maybe the log function

  6. triciaal
    • one year ago
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    right on time huh

  7. Astrophysics
    • one year ago
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    Haha :P

  8. amyna
    • one year ago
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    lol i dont understand. what you wrote, is that the correct way to do this problem or not?

  9. beginnersmind
    • one year ago
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    Nah. Start by rewriting |dw:1441505728153:dw|

  10. amyna
    • one year ago
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    oh okay so you have to use logs and rewrite it. is this true for all problems that may look similar to this?

  11. beginnersmind
    • one year ago
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    Generally base e is easier to deal with. And changing bases only changes the exponent by a constant factor. So it almost always helps.

  12. amyna
    • one year ago
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    ok thanks!

  13. amyna
    • one year ago
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    wait then how do i solve it after rewriting it? lol i forgot how to do that part!

  14. beginnersmind
    • one year ago
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    Use the chain rule.

  15. amyna
    • one year ago
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    ok thanks. i think i got it from here

  16. amyna
    • one year ago
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    no i don't get it. i don't know how to solve it. i tried using the chain rule

  17. beginnersmind
    • one year ago
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    Hm, you might need to review the chain rule then. I can go through this example if you want but I don't think it will be much help in general.

  18. beginnersmind
    • one year ago
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    @amyna

  19. amyna
    • one year ago
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    yes please! i would greatly appreciate that! :)

  20. freckles
    • one year ago
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    \[y=(f(x))^{g(x)} , \text{ assume } f(x)>0 \\ \\ \text{ take } \ln( ) \text{ of both sides } \\ \ln(y)=\ln((f(x))^{g(x)}) \\ \ln(y)=g(x) \ln(f(x)) \text{ by use of power rule for logarithms } \\ \\ \text{ now differentiate both sides } \\ \frac{y'}{y}=g'(x) \cdot \ln(f(x))+g(x) \cdot \frac{f'(x)}{f(x)} \\ \text{ by a whole bunch of rules :p } \\ \text{ left hand side I just used chain rule } \\ \text{ right hand side I used product rule and chain rule }\] \[y'=\{g'(x) \ln(f(x))+g(x) \frac{f'(x)}{f(x)} \} y \\ \text{ note: this step I just multiplied both sides by } y \] \[\text{ now remember } y=(f(x))^{g(x)} \\ \text{ so make this replacement and you are done} \\ \]

  21. beginnersmind
    • one year ago
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    Ok, I'll try to explain how to apply the chain rule. You might want to look at this lesson from MIT OCW as well: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-11-chain-rule/ You look at \(e^{(ln4)*3*x^2} \). If it was e^x the derivative would be e^x. But now the exponent is a function of x. So the result is e to that function of x _multiplied by the derivative of that function_. So \([e^{g(x)}]' = g'(x)* e^{g(x)}\) In this case g(x) = (ln4)*3*x^2 g'(x) = (ln4)*6*x so the final result is \[(ln4)*6x*e^{ln4*3*x^2} \] which you can rewrite as \[(ln4)*6x*4^{ln4*3*x^2} \] using the same idea with logarithms that we started with.

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