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Math Help - Derivative

  1. #1
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    Derivative

    I'm having trouble finding the answer to this problem. : Find the derivative of the function 1/((the square root of(x)) and the equation of the line that is tangent to the graph of f and parallel to the given line 3x-y+1. I also found the answer to this one but it was kind of guess and check. : find the equations of two tangent lines to the graph of f that pass through the indicated point where f(x)=x^2 and the indicated point is (1, -3). The last question i have is how you guys put square roots and other symbols in. Can't seem to find them. Thanks a lot.
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  2. #2
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    Quote Originally Posted by jarny View Post
    I'm having trouble finding the answer to this problem. : Find the derivative of the function 1/((the square root of(x)) and the equation of the line that is tangent to the graph of f and parallel to the given line 3x-y+1. I also found the answer to this one but it was kind of guess and check. : find the equations of two tangent lines to the graph of f that pass through the indicated point where f(x)=x^2 and the indicated point is (1, -3). The last question i have is how you guys put square roots and other symbols in. Can't seem to find them. Thanks a lot.
    The line 3x-y+1=0 can be expressed as y = 3x +1. A parallel lines means the derivative has slopt 3. Thus at which point is the derivative equal to 3?
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  3. #3
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    Don't need an answer to the first one. I made a simple algebraic mistake which messed up a rather easy problem. Thanks for the answer to the second. Rather moronic of me not to see that Man that was stupid. That was a 1st grade level mistake. Just having a bad day thinking i guess. Can anyone tell me how to make the symbols though?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jarny View Post
    find the equations of two tangent lines to the graph of f that pass through the indicated point where f(x)=x^2 and the indicated point is (1, -3).
    y = x^2

    \Rightarrow y' =2x

    so the slope of any tangent line is given by 2x

    By the point-slope form, lines passing through (1,-3) have the form:

    y + 3 = m(x - 1)

    \Rightarrow y = mx - (m + 3)

    but we want lines where the slope is given by 2x, so set m = 2x, we get:

    y = 2x^2 - (2x + 3) = 2x^2 - 2x - 3

    furthermore, we want lines that intersect with y = x^2, so equate the two functions. we get:

    x^2 = 2x^2 - 2x - 3

    \Rightarrow x = 3 or x = -1

    let m_1 be the slope of the tangent line that intersects with y = x^2 at x = 3. thus m_1 = 2x = 2(3) = 6

    so the first tangent line is: \boxed {y = 6x - 9}

    in a similar fashion, we can find the tangent line that intersects with y = x^2 at x = -1 to be \boxed {y = -2x - 1}


    The last question i have is how you guys put square roots and other symbols in. Can't seem to find them. Thanks a lot.
    Use LaTex or if you don't want to, type sqrt(x) to mean \sqrt {x}
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