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Math Help - slope of line tanget to graph

  1. #1
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    slope of line tanget to graph

    The slope of the line tangent to the graph of 3x^2 + 5ln(y) = 12 at (2,1) is...

    a) -12/5
    b) 12/5
    c) 5/12
    d) 12
    e) -7


    **Im not particularly differentiating y in respect to x and could use some help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DINOCALC09 View Post
    The slope of the line tangent to the graph of 3x^2 + 5ln(y) = 12 at (2,1) is...

    a) -12/5
    b) 12/5
    c) 5/12
    d) 12
    e) -7


    **Im not particularly differentiating y in respect to x and could use some help.
    how else would you do it?

    i'd differentiate implicitly here and then plug in x = 2, y = 1 into the formula i get.


    3x^2 + 5 \ln y = 12

    differentiating implicitly we get:

    6x + \frac 5y~\frac {dy}{dx} = 0

    now solve for dy/dx and plug in the required x- and y-values
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    how else would you do it?

    i'd differentiate implicitly here and then plug in x = 2, y = 1 into the formula i get.


    3x^2 + 5 \ln y = 12

    differentiating implicitly we get:

    6x + \frac 5y~\frac {dy}{dx} = 0

    now solve for dy/dx and plug in the required x- and y-values

    dy/dx = -6x/(5/y)?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DINOCALC09 View Post
    dy/dx = -6x/(5/y)?
    no.

    6x + \frac 5y~\frac {dy}{dx} = 0 ........move the 6x to the other side by subtracting it from both sides

    \Rightarrow \frac 5y ~\frac {dy}{dx} = -6x ........multiply both sides by \frac y5

    \Rightarrow \frac {dy}{dx} = \frac {-6xy}5
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  5. #5
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    so as im understanding, the answer is (A) or -12/5
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DINOCALC09 View Post
    so as im understanding, the answer is (A) or -12/5
    yes
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