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Math Help - About the chain rule

  1. #1
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    About the chain rule

    I have this problem that I'm going to make the derivative of:

    6000/(1+49(0.6)^t

    The next step i'm confused about:

    6000/(1+49(0.6)^t)^2 d/dx (1+49(0.6)^t)

    I think I am overlooking something here...

    How does the (1+49(0.6)^t)^2 come to be squared? Underneath the 6000.
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  2. #2
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    Re: About the chain rule

    Standard Exponent Rule, isn't it?

    For suitable values of 'a', we have \frac{d}{dx}\;x^{a} = a\cdot x^{a-1}

    a = -1 is a suitable value.
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  3. #3
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    Re: About the chain rule

    Hello, sara213!

    Differentiate: . f(x) \:=\:\frac{6000}{1+49(0.6^x)}

    The next step i'm confused about:

    f'(x) \;=\; \frac{6000}{\left[1+49(0.6^x)\right]^2}\,\frac{d}{dx}\left[1 + 49(0.6^x)\right] . not quite right

    I think I am overlooking something here . . .

    How does the (1+49\!\cdot\!0.6^t)^2 come to be squared? Underneath the 6000?

    Are you sure you know your differentiation rules?

    You can use the Quotient Rule,
    . . which ends with "over the denominator squared".


    Or use an easier way . . .

    We have: . f(x) \:=\:6000\left[1 + 49(0.6^x)\right]^{-1}

    Chain Rule: . f'(x) \;=\;6000\cdot(\text{-}1)\left[1 + 49(0.6^x)\right]^{-2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]

    . . . . . . . . . . f'(x) \;=\;\frac{-6000}{[1 + 49(0.6^x)]^2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]

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    Re: About the chain rule

    Quote Originally Posted by Soroban View Post
    Hello, sara213!


    Are you sure you know your differentiation rules?

    You can use the Quotient Rule,
    . . which ends with "over the denominator squared".


    Or use an easier way . . .

    We have: . f(x) \:=\:6000\left[1 + 49(0.6^x)\right]^{-1}

    Chain Rule: . f'(x) \;=\;6000\cdot(\text{-}1)\left[1 + 49(0.6^x)\right]^{-2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]

    . . . . . . . . . . f'(x) \;=\;\frac{-6000}{[1 + 49(0.6^x)]^2} \cdot \frac{d}{dx}\left[1 + 49(0.6^x)\right]


    I thought i did but I think the notation : d/dx is throwing me off. I like f(x) and g(x) stuff for example:

    \;=\;\frac{g(x) * f'(x) - f(x) * g'(x)}{g(x)^2}
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  5. #5
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    Re: About the chain rule

    Quote Originally Posted by sara213 View Post
    I thought i did but I think the notation : d/dx is throwing me off. I like f(x) and g(x) stuff for example:

    \;=\;\frac{g(x) * f'(x) - f(x) * g'(x)}{g(x)^2}
    If you want to use your notation then

    h(t) =\frac{6000}{1+49(0.6^t))}

    let f(t)=6000 and g(t)=(1+46(0.6^t))

    h'(t)=\frac{-6000* \frac{d}{dt} (1+49(0.6^t))}{(1+49(0.6^t))^2}
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  6. #6
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    Re: About the chain rule

    When your numerator is a constant, as here, it is perhaps easier to think of the problem as having a negative exponent:
    \frac{6000}{1+ 49(.06)^t}= 6000(1+ 49(.06)^t)^{-1}

    Now, that the derivative is 6000(-1)(1+ 49(.06)^t)^{-2}(d/dx(1+ 49(.06)^t)

    So you can think of the denominator being (1+ 49(.06)^t) being squared as a result of either the quotient rule or simply the power rule: (x^n)'= nx^{n-1} with n= -1 so that n-1= -1-1= -2.

    Note that to finish you will need the fact that the derivative of a^t is a^t(ln(a)).
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  7. #7
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    Re: About the chain rule

    Quote Originally Posted by sara213 View Post
    I have this problem that I'm going to make the derivative of:
    6000/(1+49(0.6)^{\color{red}t}

    The next step i'm confused about:
    6000/(1+49(0.6)^t)^2 d/\color{red}dx (1+49(0.6)^t)
    \text{Is the variable }t\text{ or }x~?
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  8. #8
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    Re: About the chain rule

    Quote Originally Posted by Plato View Post
    \text{Is the variable }t\text{ or }x~?
    It would be t
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