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Math Help - PLease Differentiate...

  1. #1
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    PLease Differentiate...

    1.)y=x^5(e^{-3lnx })
    The answer for this is 2x

    2.) FInd an equation of the normal line to the curve y = \sqrt{x^2 + 9} at the point (4,5)

    Thanks
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)y=x^5(e^{-3lnx })
    The answer for this is 2x
    Note sure what this first question is asking.

    2.) FInd an equation of the normal line to the curve y = \sqrt{x^2 + 9} at the point (4,5)
    The slope of the curve y = \sqrt{x^2 + 9} at xis:

    m(x)=\frac{d}{dx}\sqrt{x^2 + 9}=\frac{x}{\sqrt{x^2+9}}

    so when x=4 the slope is:

    m(4)=\frac{4}{\sqrt{16+9}}=4/5.

    The normal to the curve at (4,5) has slope -1/m(4)=-5/4, and passess through the point, so the equation on the normal is:

    y=-(5/4)x+c

    where:

    5=-(5/4)5+c=5+c, so c=10, and the equation is:

    y=-(5/4)x+10

    RonL
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)y=x^5(e^{-3lnx })
    The answer for this is 2x
    I think he wants help differentiating it.

    y = x^5*e^(-3*ln(x))

    dy/dx = 5x^4*e^(-3*ln(x)) + x^5*(deriv. of e^(-3*ln(x)) here)

    -3*1/x*e^(-3*ln(x)) = -3/x*e^(-3*ln(x)) = (-3/x)*(1/x^3) = -3/x^2

    Thus, we have:

    5x^4*e^(-3*ln(x)) + x^5*(-3/x^2)

    5x^4*e^(-3*ln(x)) - 3x

    5x^4*1/(x^3) - 3x

    (5x^4)/(x^3) - 3x

    5x - 3x = 2x, which agrees with the answer you thought it might be.
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  4. #4
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)y=x^5(e^{-3lnx })
    The answer for this is 2x

    2.) FInd an equation of the normal line to the curve y = \sqrt{x^2 + 9} at the point (4,5)

    Thanks
    Hello,

    to 1.)

    Simplify first the term

    e^{-3 \cdot \ln(x)}=e^{ \ln(x^{-3})}=e^{ \ln\left(\frac{1}{x^3} \right)}=\frac{1}{x^3}. Now your function becomes:

    y=x^5 \cdot \frac{1}{x^3}=x^2

    I leave the derivation for you

    EB
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  5. #5
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1.)y=x^5(e^{-3lnx })
    The answer for this is 2x

    2.) FInd an equation of the normal line to the curve y = \sqrt{x^2 + 9} at the point (4,5)

    Thanks
    Here is one way.

    1) y = (x^5)[e^(-3ln(x))]
    Differentiate both sides with respect to x.

    Let us evaluate first the e^(-3ln(x)).
    Let e^(-3ln(x)) = N
    Take the ln of both sides,
    (-3ln(x))ln(e) = ln(N)
    (-3ln(x))(1) = ln(N)
    -3ln(x) = ln(N)
    ln(x^(-3)) = ln(N)
    Hence,
    N = x^(-3)
    So, e^(-3ln(x)) = x^(-3) --------***

    That's very long, I know, but I don't know if you know that
    e^(ln x^(-3)) = x^(-3).

    So,
    y = (x^5)[e^(-3ln(x))]
    y = (x^5)(x^(-3))
    y = x^2
    dy/dx = 2x --------------answer.

    ----------------------------------------------------
    2) 2.) FInd an equation of the normal line to the curve y = sqrt(x^2 +9) at the point (4,5)

    dy/dx is the slope of the tangent line.
    The slope of the normal line is the negative reciprocal dy/dx, which is -dx/dy.

    y = sqrt(x^2 +9)
    Square both sides,
    y^2 = x^2 +9
    x = sqrt(y^2 -9)
    dx/dy = (1/2)[(y^2 -9)^(-1/2)](2y)
    dx/dy = y /sqrt(y^2 -9)
    At point (4,5),
    dx/dy = 5 /sqrt(5^2 -9) = 5 /sqrt(16) = 5/4
    Hence, slope of normal line = -5/4.

    Slope = -5/4
    Point(4,5)
    Point-slope form of the equation of the line,
    (y -y1) = m(x -x1)
    (y -5) = (-5/4)(x -4)
    Clear the fraction, multiply both sides by 4,
    4y -20 = -5x +20
    5x +4y -40 = 0 -------------------the normal line, in standard form

    In y = mx +b form,
    5x +4y -40 = 0
    4y = -5x +40
    y = -(5/4)x +10 -------------the normal line.
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