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Thread: using chain rule to solve an equation

  1. #1
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    using chain rule to solve an equation

    Hi

    I am struggling to solve the equations:

    Given:

    d(e^t)/dt =e^t

    use the chain rule to find dy/dx:
    a) y=e^5t,
    b) y=4e^3t
    c)y=6e^-2t

    (where "^" represents an exponent)
    Thanks in advance for any help

    Motty
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  2. #2
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    $\displaystyle y=e^{5t}=\{u=5t\}=e^{u(t)}$

    $\displaystyle \frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}=e^u\frac{ du}{dt} = e^{u}\cdot 5 = 5e^{5t}$

    Use the same process for the other questions.
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  3. #3
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    The Chain Rule states,

    $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x).$

    It can be seen to work through the fact that $\displaystyle g(x)$ changes at the rate $\displaystyle g'(x)$ as $\displaystyle x$ moves, speeding up (or slowing down) the rate of change of $\displaystyle f(g(x))$ by that amount.

    For instance, if $\displaystyle g'(x)=2$, then $\displaystyle g(x)$ moves twice as fast, and $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot 2$.

    For problem (a), we note that $\displaystyle f(t)=e^t$ and $\displaystyle g(t)=5t$. The Chain Rule therefore states,

    $\displaystyle \frac{d}{dt}(e^{5t})=\frac{d}{dt}f(g(t))=f'(g(t))g '(t)=e^{g(t)}\frac{d}{dt}(5t)=e^{5t}\cdot 5=5e^{5t}.$

    The same reasoning may be applied to (b) and (c).
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  4. #4
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    Thanks guys, following the same reasonig, I think, does that mean:
    b) = 12e^3t
    c) = -12e^-2t
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