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Math Help - Derivative of Exponential Function Questions

  1. #1
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    Derivative of Exponential Function Questions

    "For the following, determine the equation of the tangent at the given point:
    12b) For the curve defined by (x^2)(e^y)=1 at B(1,0)."

    "The number, N, of bacteria in a culture at time t in hours is:
    N = 1000[30 + (e^-t/30)]. Find the rate of change of the number of bacteria at time t."

    Two different questions, two question I can't solve. Any insight would be appreciated, thanks!
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  2. #2
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    Quote Originally Posted by Jeavus View Post
    "For the following, determine the equation of the tangent at the given point:
    12b) For the curve defined by (x^2)(e^y)=1 at B(1,0)."

    "The number, N, of bacteria in a culture at time t in hours is:
    N = 1000[30 + (e^-t/30)]. Find the rate of change of the number of bacteria at time t."

    Two different questions, two question I can't solve. Any insight would be appreciated, thanks!
    For the first one, you need to use Implicit Differentiation. You can start by reshaping the equation so it reads
    x^2=e^{-y} When you differentiate both sides, you should get 2x\delta x=-e^{-y} \delta y

    Again, you can reorder this one more time so you have \frac{\delta y}{\delta x}=[-2xe^y]

    At this point, plug in the point you're given so you get \frac{\delta y}{\delta x}=-2

    Now you can find the equation for the line using the slope and point you have.

    (y-0)=-2(x-1)

    For the second problem, just use Differentiation to get \frac{\delta N}{\delta x}

    Try this one on your own. I will help you if you need it.
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  3. #3
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    Quote Originally Posted by Jeavus View Post
    "For the following, determine the equation of the tangent at the given point:
    12b) For the curve defined by (x^2)(e^y)=1 at B(1,0)."

    "The number, N, of bacteria in a culture at time t in hours is:
    N = 1000[30 + (e^-t/30)]. Find the rate of change of the number of bacteria at time t."

    Two different questions, two question I can't solve. Any insight would be appreciated, thanks!
    for the first one isolate y first

    x^2e^y=1 \iff e^y=\frac{1}{x^2} \iff y=\ln\left( \frac{1}{x^2}\right) \iff y =\ln(x^{-2}) \iff y=-2\ln(x)

    Now taking the derivative we get

    \frac{dy}{dx}=\frac{-2}{x} eval at x=1 and we get

    \frac{dy}{dx}|_{x=1}=\frac{-2}{1}=-2

    Then we get

    y-0=-2(x-1) \iff y=-2x+2

     <br />
N = 1000[30 + (\frac{e^{-t}}{30})]=3000 + \frac{100}{3}e^{-t}.<br />

    taking the derivative(is the rate of change) we get

    \frac{dN}{dt}=-\frac{100}{3}e^{-t}
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    for the first one isolate y first

    x^2e^y=1 \iff e^y=\frac{1}{x^2} \iff y=\ln\left( \frac{1}{x^2}\right) \iff y =\ln(x^{-2}) \iff y=-2\ln(x)

    Now taking the derivative we get

    \frac{dy}{dx}=\frac{-2}{x} eval at x=1 and we get

    \frac{dy}{dx}|_{x=1}=\frac{-2}{1}=-2

    Then we get

    y-0=-2(x-1) \iff y=-2x+2

     <br />
N = 1000[30 + (\frac{e^{-t}}{30})]=3000 + \frac{100}{3}e^{-t}.<br />

    taking the derivative(is the rate of change) we get

    \frac{dN}{dt}=-\frac{100}{3}e^{-t}
    Haha...or you can just look at the answer that he gave you. lol
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