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Math Help - Double check these derivatives

  1. #1
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    Double check these derivatives

    Clearly I'm missing at least one important step.

    (1) Find an equation of the line tangent to the curve y1 given below that is parallel to the given line y2.

    y
    1=6x * sqrt(x)
    y2 = 9 + 9x

    y1 = 6x(x^.5)
    y1 = 6x^(3/2)
    y1' = (3/2)(6x^.5)
    y1' = 9x^(1/2)

    Seems to match the criteria, but it's wrong?

    (2) For what values of r does the function y = erx satisfy the equation y'' + 3y' - 40y = 0?

    (3) f (x) = 9 cos(x) + 5/7 cot(x)

    Seemed simple.

    9(-sin(x)) + (5/7)(-csc^2(x))

    (4) 6-6tan(x) /sin(x)-cos(x)

    limit as x goes to pi/4. Bottom is 0 = ??

    (5) Implicit differentiation cos(x) + sqrt(y) = 7

    -sin(x) + (1/2)y^(-1/2) * (dy/dx) = 0
    (1/2)y^(-1/2) = sin(x)
    y^(-1/2) = 2sin(x)
    1/sqrt(y) = 2sin(x)
    y = (1/2sin(x))^2
    y = 1/4sin(x)^2

    Again, not sure where I went wrong.

    (6) 5sqrt(x) + sqrt(y) = 5
    (5/2)x^(-1/2) + (1/2)y^(-1/2) * (dy/dx) = 0

    I can't put this answer into a form that doesn't suck, but it was 5/sqrt(x) over 1/sqrt(y)

    (7) f(x) = sin(4x) + ln(4x)
    f'(x) = cos(4x) + 1/4x

    This one seems simple, but I'm guessing the problem is with the Trig part.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by CFem View Post
    Clearly I'm missing at least one important step.

    (1) Find an equation of the line tangent to the curve y1 given below that is parallel to the given line y2.

    y
    1=6x * sqrt(x)
    y2 = 9 + 9x

    y1 = 6x(x^.5)
    y1 = 6x^(3/2)
    y1' = (3/2)(6x^.5)
    y1' = 9x^(1/2)

    Seems to match the criteria, but it's wrong?
    y = 6x\sqrt x  = 6\sqrt {{x^3}}  \Rightarrow {y^\prime } = 9\sqrt x .

    The equation of the line tangent:

    y - 6\sqrt {x_0^3}  = 9\sqrt {{x_0}} \left( {x - {x_0}} \right)

    Now use the condition of parallelism k_1=k_2 of two straight lines y_1= k_1 x + a, {y_2} = {k_2}x + b.

    Then you get: 9 = 9\sqrt{x_0}\Leftrightarrow x_0 = 1.

    Finally, you have: y - 6 = 9\left( {x - 1} \right) \Leftrightarrow y = 9x - 3.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by CFem View Post
    Clearly I'm missing at least one important step.

    (4) 6-6tan(x) /sin(x)-cos(x)

    limit as x goes to pi/4. Bottom is 0 = ??

    This one seems simple, but I'm guessing the problem is with the Trig part.
    \mathop {\lim }\limits_{x \to \pi /4} \frac{{6 - 6\tan x}}{{\sin x - \cos x}} = 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{1 - {{\tan }^2}x}}{{\left( {\sin x - \cos x} \right)\left( {1 + \tan x} \right)}} =

    = 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{{{\cos }^2}x -{{\sin }^2}x}}{{{{\cos }^2}x\left( {\sin x - \cos x} \right)\left( {1 + \tan x} \right)}} =

    = 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}{{{{\cos }^2}x\left( {\sin x - \cos x} \right)\left( {1 + \tan x} \right)}} =

    =  - 6\mathop {\lim }\limits_{x \to \pi /4} \frac{{\cos x + \sin x}}{{{{\cos }^2}x\left( {1 + \tan x} \right)}} =

    =  - 6 \cdot \frac{{\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}}}<br />
{{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}\left( {1 + 1} \right)}} =  - 6\sqrt 2 .
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