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I Derivative of 4^x

  1. Apr 25, 2017 #1
    On my exam, we had to find the derivative of 4^x. This is what I did
    Y=4^x
    lny=xln4
    y=e^xln4
    and then finding the derivative for that I got, (xe^(xln4))/4
    My professor said that it was wrong and even after I told her what I did to get the answer. She told me the answer was (4^x)ln4 . Which I know it is but I think this is still equivalent to my answer. Was I right? Regardless I still don't get the point :(
     
  2. jcsd
  3. Apr 25, 2017 #2

    Dr Transport

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    take the derivative of [itex] \ln(y)[/itex] and then substitute [itex] y [/itex] back into the result to get the professors answer.....
     
  4. Apr 25, 2017 #3
    How did you get that as the derivative?
     
  5. Apr 25, 2017 #4
    I did the chain rule, so I got x*1/4*e^xln4 . Which I now realize is wrong, it should've been x*1/4+ln4*e^xln4. But was I right with the y=e^xln4?
     
  6. Apr 25, 2017 #5
    You're still doing it wrong but yes ##y=e^{xln4}## is correct
     
  7. Apr 26, 2017 #6

    Dr Transport

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    $$y = 4^x$$

    $$ \ln(y) = \ln(4^x) = x\ln(4)$$

    $$\frac{dy}{y} = dx \ln(4)$$

    $$\frac{dy}{dx} = y \ln(4) = x^4 \ln(4)$$

    proves the instructors answer and no need in taking exponentials.....
     
  8. Apr 27, 2017 #7
    You mean to say that the final answer is:
    dy/dx = yln(4) = (4^x)ln(4)
    Right?
     
  9. Apr 27, 2017 #8
    Also, why is derivative of ln(4) not evaluated as 1/4?
     
  10. Apr 28, 2017 #9

    haushofer

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    Because the derivative of a constant is zero.
     
  11. Apr 28, 2017 #10

    Dr Transport

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    yes, typo......
     
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