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Math Help - Differentiation

  1. #1
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    Differentiation

    1. If d/dx(f(x)) = g(x) and if h(x) = x^2, then d/dx(f(h(x)) = ?
    A) g(x^2)
    B) 2xg(x)
    C) g'(x)
    D) 2xg(x^2)
    E) x^2g(x^2)

    I got A,
    d/dx(f(h(x))
    = g(h(x))
    = g(x^2)

    Not 100% sure it that's right so I'd like to get someone to check it please.


    2.

    In the figure, AB is a 40 ft ladder with end A against a vertical wall and end B on level ground. If the ladder is sliding down the wall, what is the distance BC at the instant when B is moving along the ground 3/4 as fast as A is moving down the wall?

    I derived pythagoreans theorem to get
    2x(dx/dt) + 2y (dy/dt) = 0
    The problem is I don't have values for either x or y and I'm not sure how to get them.
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  2. #2
    Super Member redsoxfan325's Avatar
    Joined
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    Swampscott, MA
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    Quote Originally Posted by Naples View Post
    1. If d/dx(f(x)) = g(x) and if h(x) = x^2, then d/dx(f(h(x)) = ?
    A) g(x^2)
    B) 2xg(x)
    C) g'(x)
    D) 2xg(x^2)
    E) x^2g(x^2)

    I got A,
    d/dx(f(h(x))
    = g(h(x))
    = g(x^2)

    Not 100% sure it that's right so I'd like to get someone to check it please.


    2.

    In the figure, AB is a 40 ft ladder with end A against a vertical wall and end B on level ground. If the ladder is sliding down the wall, what is the distance BC at the instant when B is moving along the ground 3/4 as fast as A is moving down the wall?

    I derived pythagoreans theorem to get
    2x(dx/dt) + 2y (dy/dt) = 0
    The problem is I don't have values for either x or y and I'm not sure how to get them.
    Number one is the chain rule: \frac{d}{dx}\left[f(h(x))\right] = f'(h(x))\cdot h'(x), which in this case is g(x^2)\cdot 2x, so the answer is D.

    ---------------

    For number two: Let x=CB and y=AB. We want to solve the equation dx=\frac{3}{4}dy for x.

    We know that x^2+y^2=40 so y=\sqrt{40-x^2} and dy=\frac{-x}{\sqrt{40-x^2}}\,dx

    We can ignore the minus sign because that just indicates direction and does not affect magnitude. Now we need to solve dx=\frac{3}{4}\cdot\frac{x}{\sqrt{40-x^2}}\,dx.

    Canceling the dx's and cross-multiplying gives us 4\sqrt{40-x^2}=3x \implies 16(40-x^2)=9x^2 \implies 25x^2=640 \implies x=\sqrt{\frac{124}{5}}=\frac{2\sqrt{155}}{5}
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