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Math Help - Differenciate

  1. #1
    Member srirahulan's Avatar
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    Lightbulb Differenciate

    How can i differenciate e^\frac{y}{x}=x respect to x>>>>>
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  2. #2
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    Re: Differenciate

    Pretty straightforward, using the "chain rule". First, of course, we integrate functions, not equations, so what you really mean is "how do I differentiate each side of this equation?"

    To differentiate e^{\frac{y}{x}}, let u=\frac{y}{x}= yx^{-1}. Then \frac{de^u}{dx}= \frac{de^u}{du}\frac{du}{dx}. \frac{de^u}{du}= e^u and \frac{dyx^{-1}}{dx}= (-1)yx^{-2}= -\frac{y}{x^2}. So the derivative of e^{\frac{y}{x}}, with respect to x is -\frac{y}{x^2}e^{\frac{y}{x}}.

    Now, what is the derivative of the right side of the equation?
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  3. #3
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    Re: Differenciate

    Hello, srirahulan!

    \text{Differentiate: }\:e^\frac{y}{x}\:=\:x

    \text{Take logs: }\:\ln\left(e^{\frac{y}{x}}\right) \:=\:\ln(x) \quad\Rightarrow\quad \frac{y}{x}\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(x)

    . . . . . . . . \frac{y}{x} \:=\:\ln(x) \quad\Rightarrow\quad y \:=\:x\!\cdot\!\ln(x)

    \text{Differentiate: }\:y' \;=\;x\!\cdot\!\frac{1}{x} + 1\!\cdot\!\ln(x)

    . . \text{Therefore: }\:y' \;=\;1 + \ln(x)
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  4. #4
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    Re: Differenciate

    Differenciate-09-aug-13-2.png
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