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Math Help - Differentiation

  1. #1
    No one in Particular VonNemo19's Avatar
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    Differentiation

    Differentiate

    y=(cosx)^{3x} with respect to x.


    Using the basic differentiation rules for elementary functions, how do I approach this? What can be defined as u?

    Is it (cosx)^u or u^{3x} ?
    Last edited by VonNemo19; June 21st 2009 at 09:58 AM.
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Differentiate

    y=(cosx)^{3x} with respect to x.


    Using the basic differentiation rules for elementary functions, how do I approach this? What can be defined as u?

    Is it (cosx)^u or u^{3x} ?
    Logarithmic differentiation will do here.

    ln y = ln {(cos {x})^{3x}}

    ln y = 3x ln {cos{x}}

    Now implicitly differentiate, then isolate \frac{dy}{dx}

    I hope this helps.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    That's what I thought, but I've never applied logs to trig functions before, so i was hesitant. Thanx.




    Difeerentiating

    \frac{d}{dx}(lny)=\frac{d}{dx}(3xlncosx)

    \frac{y'}{y}=3x\frac{-sinx}{cosx}+3lncosx

    y=3xlncosx

    y'=9xlncosx(\frac{-xsinx}{cosx}+lncosx)

    Is this correct?
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    That's what I thought, but I've never applied logs to trig functions before, so i was hesitant. Thanx.




    Difeerentiating

    \frac{d}{dx}(lny)=\frac{d}{dx}(3xlncosx)

    \frac{y'}{y}=3x\frac{-sinx}{cosx}+3lncosx

    y=3xlncosx

    y'=9xlncosx(\frac{-xsinx}{cosx}+lncosx)

    Is this correct?
    No. I have marked in red what is uncorrect.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by running-gag View Post
    No. I have marked in red what is uncorrect.

    Duh! Unbelievable. Thanx
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  6. #6
    Super Member Random Variable's Avatar
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    Let u = cos(x) and v = 3x

    then differentiate  y= u^{v} by using the chain rule for functions of more than one variable

     \frac {dy}{dx} = \frac {\partial y}{\partial u} \ \frac {du}{dx} + \frac {\partial y}{\partial v} \ \frac {dv}{dx}
    Last edited by Random Variable; June 21st 2009 at 12:08 PM.
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