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Math Help - Implicit Differentiation

  1. #1
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    Implicit Differentiation

    Here is the problem:
    Use implicit differentiation to show (prove) that the derivative of y(x)=log 22 (x) is y'(x) = 22x ln (22).

    Here is what I have done so far:

    y(x) = (lnx)/(ln 22)

    Differentiating,
    y'(x) = 1/(x ln 22).

    but I do not get what is asked for.
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  2. #2
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    Re: Implicit Differentiation

    Hey mymath.

    Your y'(x) is definitely not going to be that: I think the question is asking you to find the derivative of y(x) = 22^(x) instead of y(x) = log_22(x).
    Thanks from mymath
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  3. #3
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    Re: Implicit Differentiation

    Hello, mymath!

    I agree with Chiro.
    I suspect that the problem reads like this:


    \text{Using implicit differentiation,}
    \text{prove that the derivative of }\,{\color{red}y \:=\:22^x}\,\text{ is }\,y' \:=\: 22^x\ln(22)

    We have: . y \;=\:22^x

    Take logs: . \ln(y) \:=\:\ln(22^x) \quad\Rightarrow\quad \ln(y) \:=\:x\cdot\ln(22)

    Differentiate: . \frac{1}{y}\!\cdot\!y' \:=\:1\cdot\ln(22)

    Therefore: . y' \:=\:y\cdot\ln(22) \quad\Rightarrow\quad y' \:=\:22^x\ln(22)

    Thanks from mymath
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  4. #4
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    Re: Implicit Differentiation

    yes, Soroban, you are right.

    y(x) = 22^x.

    Thanks a lot.
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  5. #5
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    Re: Implicit Differentiation

    Chiro, you are right.

    y(x) = 22^x.

    Thank you.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Implicit Differentiation

    A similar approach would be to write:

    y=22^x=e^{\ln(22^x)}=e^{x\ln(22)}

    Now, differentiate:

    \frac{dy}{dx}=e^{x\ln(22)}\ln(22)=\ln(22)22^x
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