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Math Help - Calculus boggles my brains :[

  1. #1
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    Exclamation Calculus boggles my brains :[

    Hey, can you guys explain how to do these problems? I'm stumped!

    1) If f(x)=sinx and g(x)= (x+2)^(1/2) then the derivative of y=g(f(x)) at x=pi is ...

    2) The lim as h -> 0 ( tan 4(x+h) - tan(4x) )/ h is....

    3) Find the derivative of y=(3x^4)e^3x

    4) If ln (4x) = e^y, what is dy/dx in terms of x?

    5) If y = ln x, then the third derivative is...

    7) If y = (g(x))^3 find y'(2) when g(2)=2 and g'(2)=10

    8) Consider the curve defined by xy^2-x^3y=6
    Find all points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by NoiseandAttack View Post
    1) If f(x)=sinx and g(x)= (x+2)^(1/2) then the derivative of y=g(f(x)) at x=pi is ...
    If y=g[f(x)]

    differentiate
    y\prime=\frac{d}{dx}g[f(x)]

    So differentiate g(x) and apply the chain rule
    y\prime=g\prime[f(x)]f\prime(x)

    So just find g'(x) and f'(x), then plug f(x) into g'(x) and multiply that by f'(x)
    Last edited by angel.white; December 5th 2007 at 05:42 PM.
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  3. #3
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    Quote Originally Posted by angel.white View Post
    If y=g[f(x)]

    differentiate
    y\prime=\frac{d}{dx}g[f(x)]

    So differentiate g(x) and apply the chain rule
    y\prime=g\prime[f(x)]\prime(x)

    So just find g'(x) and f'(x), then plug f(x) into g'(x) and multiply that by f'(x)
    I still don't seem to be getting the right answer

    my work:
    g'(x) = 1/2(x+2)^-.5
    = (x+2)/ sqrt of 2
    f'(x) = cosx

    (cos x) + 2 / sqrt of 2 * cos x

    cos pi +2 / sqrt of 2 * cos pi
    -1/ sqrt of 2
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  4. #4
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    error

    NoiseandAttack,

    g'(x)=\tfrac{1}{2}(x+2)^{-\tfrac{1}{2}}
    which equals \frac{1}{2 \sqrt{x+2}}, not \frac{x+2}{\sqrt{2}}.

    --Kevin C.
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  5. #5
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    Error

    Further, it's y'=g'(f(x))f'(x), so you should be putting f(x)=\sin x into g', not f'(x)=\cos x.

    --Kevin C.
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  6. #6
    Senior Member polymerase's Avatar
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    f(x)=(sin\:x +2)^{\frac{1}{2}}
    f'(x)=\frac{1}{2}(sin\:x+2)^{-\frac{1}{2}}(cos\:x)
    f'(\pi)=\frac{1}{2}(0+2)^{-\frac{1}{2}}(-1)
    f'(\pi)=-\frac{1}{2\sqrt{2}}

    y=(3x^4)e^{3x}
    y'=(12x^3)(e^{3x})+(3e^{3x})(3x^4)
    y'=3e^{3x}x^3(3x+4)

    ln(4x)=e^y to get this in terms of x, you have to ln both side
    So... y=ln(ln(4x))
    ln(4x)=e^y ...use implict to find y'
    \frac{4}{4x}=e^y\frac{dy}{dx}
    \dfrac{dy}{dx}=\frac{1}{xe^y} = \frac{1}{xe^{ln(ln(4x))}}

    y=ln\:x log differentiation
    y'=\frac{1}{x} quotient rule
    y''=-\frac{1}{x^2} quotient rule
    y'''=\frac{2}{x^3}

    f(x)=(g(x))^3
    f'(x)=3(g(x))^2\:g'(x)
    f'(2)=3(g(2))^2\:g'(2)
    f'(2)=3(2)^2(10)
    f'(2)=120
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