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Math Help - Chain rule

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    Chain rule

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    Quote Originally Posted by Mathmania View Post
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    y = \frac{\sqrt{3x^2+4}}{(x^2-3)^3}

    Recall that if we have y = f(g(x)) that
    y' = f'(g(x)) \cdot g'(x)

    And if y = f(g(h(x)))
    y' = f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x)

    etc, etc.

    So to facilitate this process a bit I'm going to rewrite this as:
    y = \sqrt{3x^2+4}(x^2-3)^{-3}

    We are going to be using the product rule: y = f(x)g(x) --> y' = f'(x)g(x) + f(x)g'(x). Then we use the chain rule on the individual factors.

    y' = \left ( \frac{\frac{1}{2}}{\sqrt{3x^2+4)}} \cdot (6x) \right ) (x^2 - 3)^{-3} + \sqrt{3x^2+4} \cdot \left ( (-3) \cdot (x^2-3)^{-4} \cdot (2x) \right )

    After a bit of simplifying I get:
    y' = \frac{-3x(5x^2+11)\sqrt{3x^2+4}}{(x^2-3)^4(3x^2+4)}

    Now. At the point (2, 4) the slope has a value of -93/2. Thus the tangent line will be:
    y = -\frac{93}{2}x + b
    and will pass through the point (2, 4):

    4 = -\frac{93}{2} \cdot 2 + b

    4 = -93 + b

    b = 97

    Thus the tangent line is
    y = -\frac{93}{2}x + 97

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathmania View Post
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    3. g(x) = \sqrt{(4x+5)^3 - 1}

    g'(x) = \frac{ \frac{1}{2} }{\sqrt{(4x+5)^3 - 1}} \cdot 3(4x+5)^2 \cdot 4

    g'(x) = \frac{6(4x+5)^2}{\sqrt{(4x+5)^3 - 1}}

    -Dan
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    Quote Originally Posted by Mathmania View Post
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    Like the first problem we need to have the value of the slope of the function. So:
    y = \frac{(x^2 + 1)^3}{(x^2 - 1)^3}

    I'll use the quotient rule here, then the chain rule for each factor:
    y' = \frac{ (3(x^2+1)^2 \cdot (2x))(x^2 - 1)^3 - (x^2+1)^3(3(x^2 - 1)^2 \cdot (2x)) }{\left ( (x - 1)^3 \right ) ^2}

    After a bit of simplification:
    y' = - \frac{12x(x + 1)^2}{(x - 1)^4}

    Now for the first tangent line. We want the tangent at the point \left ( -3, \frac{1000}{512} \right ). The slope will be:
    y' = - \frac{12(-3)((-3)^2 + 1)^2}{((-3)^2 - 1)^4} = \frac{225}{256}

    So the first line has the form y = \frac{225}{256}x + b. Inserting the point \left ( -3, \frac{1000}{512} \right ) into this I get that b = \frac{1175}{256}.

    So the first line is: y = \frac{225}{256}x + \frac{1175}{256}.

    In a similar fashion the line tangent to the function at the point \left ( -2, \frac{125}{27} \right ) is: y = \frac{200}{27}x + \frac{175}{9}.

    To find the point of intersection we need to solve the system of equations:
    y = \frac{225}{256}x + \frac{1175}{256}

    y = \frac{200}{27}x + \frac{175}{9}

    Use your favorite method here. It's ugly and if you want help I'll provide it. Until then I'll simply state the solution:
    The point of intersection is: \left ( - \frac{4107}{1805}, \frac{935}{361} \right ).

    Two comments:
    1) Check my math. This is sufficiently complicated I may have made a small error somewhere. I'd simply graph the problem, but my nicer graphing software isn't available. (On my TI-92 it seems to check out, but the resolution is bad.)

    2) Whoever came up with this one is a sadist!

    -Dan
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    Hello, Mathmania!

    1) Find the equation of the tangent line to the curve

    . . y \:=\:\frac{\sqrt{3x^2+4}}{(x^2-3)^3} at the point (2,4).

    I would use Log Differentiation . . .

    We have: . \ln y \;=\;\frac{1}{2}\ln(3x^2+4) - 3\ln(x^2-3)

    . . Then: . \frac{1}{y}\!\cdot\!y' \;=\;\frac{1}{2}\!\cdot\!\frac{6x}{3x^2+4} - 3\!\cdot\!\frac{2x}{x^3-3} \quad\Rightarrow\quad y' \;=\;y\left[\frac{3x}{3x^2+4} - \frac{6x}{x^2-3}\right]

    At (2,4):\;y' \;=\;4\left[\frac{6}{16} - \frac{12}{1}\right] \;=\;4\left(-\frac{93}{8}\right)\;=\;-\frac{93}{2}

    We have: . y - 4 \:=\:-\frac{93}{2}(x - 2)\quad\Rightarrow\quad\boxed{y \;=\;-\frac{93}{2}x + 97}



    Given: f(x)\:=\:4x+5 and g(x) \:=\:\sqrt{\left[f(x)\right]^3 - 1}
    . . Determine: . g'(x).

    We have: . g(x) \;=\;\left[(4x+5)^3 - 1\right]^{\frac{1}{2}}


    Chain Rule: . g'(x) \;=\;\frac{1}{2}\left[(4x+5)^3-1\right]^{-\frac{1}{2}}\cdot3(4x+5)^2\cdot4

    Therefore: . \boxed{g'(x)\;=\;\frac{6(4x+5)^2}{\sqrt{(4x+5)^3-1}}}

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    for the last part in question 2 can u show me what method you used, i tried using 2 different methods but i get a weird answer
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathmania View Post
    for the last part can u show me what method you used, i tried using 2 different methods but i get a weird answer

    nevermind got it
    I'm presuming this message is directed at me. I'm going to post the solution anyway, just in case someone wants it.

    y = \frac{225}{256}x + \frac{1175}{256}

    y = \frac{200}{27}x + \frac{175}{9}

    Note: A graphing calculator would be very handy here. If you don't have one I would recommend using decimals instead of trying to combine the fractions by hand. (This is one of the rare places I suggest that.)

    We are looking for a place where x and y are the same for both equations. Since both are solved for y, we can say that the right hand sides of both equations are equal:
    \frac{225}{256}x + \frac{1175}{256} = \frac{200}{27}x + \frac{175}{9}

    \left ( \frac{225}{256} - \frac{200}{27} \right ) x = \frac{175}{9} - \frac{1175}{256}

    - \frac{45125}{6912} x = \frac{34225}{2304}

    x = - \frac{34225}{2304} \cdot \frac{6912}{45125} = - \frac{4107}{1805}

    Then use either of the original equations to find y. I'll use the first:
    y = - \frac{225}{256} \cdot \frac{4107}{1805} + \frac{1175}{256}

    y = - \frac{184815}{92416} + \frac{1175}{256} = \frac{935}{361}

    -Dan
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