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Thread: 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 $\displaystyle y=g[f(x)]$

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

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

    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; Dec 5th 2007 at 04:42 PM.
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  3. #3
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    Quote Originally Posted by angel.white View Post
    If $\displaystyle y=g[f(x)]$

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

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

    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,

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

    --Kevin C.
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  5. #5
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    Error

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

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

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

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

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

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