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Math Help - Derivatives and tangent lines

  1. #1
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    Derivatives and tangent lines

    I'm having some trouble on these questions. On other questions for this I was using derivatives to get the answers but on these I don't know how to apply it. Can someone explain how to use derivatives to find these answers?

    1. Find the equation of the tangent line to the curve given by y=2xe^x. At (0,0).


    2. Find the exact points (x,y) on the curve y=x ln x where the tangent line is horizontal.

    3. Find the exact point (x,y) on the curve y= sin^2 x, 0<x<pie where the tangent is horizontal.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by goldenroll View Post
    I'm having some trouble on these questions. On other questions for this I was using derivatives to get the answers but on these I don't know how to apply it. Can someone explain how to use derivatives to find these answers?

    1. Find the equation of the tangent line to the curve given by y=2xe^x. At (0,0).


    2. Find the exact points (x,y) on the curve y=x ln x where the tangent line is horizontal.

    3. Find the exact point (x,y) on the curve y= sin^2 x, 0<x<pie where the tangent is horizontal.
    1. y'=2xe^x+2e^x\Rightarrow{m}=2 at x=0.
    Then the line tangent at (0,0) is given by (y-0)=2(x-0).

    2. Horizontal \Rightarrow{y'}=0. Then y'=1+\ln{x}=0\Rightarrow{x}=\frac{1}{e}

    Write the line in the same manner as the first problem I laid out.

    3. I leave this to you
    Last edited by VonNemo19; July 27th 2009 at 08:49 PM. Reason: Made a horrible mistake... I've been doing this a lot lately...
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  3. #3
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    Quote Originally Posted by goldenroll View Post
    I'm having some trouble on these questions. On other questions for this I was using derivatives to get the answers but on these I don't know how to apply it. Can someone explain how to use derivatives to find these answers?

    1. Find the equation of the tangent line to the curve given by y=2xe^x. At (0,0).
    The derivative is the slope of the tangent line. The tangent line at (0,0) is y= m(x- 0)+ 0= mx where m is the derivative. What is the derivative of y= 2xe^x? What is that when x= 0?


    2. Find the exact points (x,y) on the curve y=x ln x where the tangent line is horizontal.
    A line is horizontal when its slope is 0. What is the derivative of x ln x? Set that equal to 0 and solve for x.

    3. Find the exact point (x,y) on the curve y= sin^2 x, 0<x<pie where the tangent is horizontal.
    Same as (2).
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  4. #4
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    Oh so on the first one am i using the chain rule? since its 2xe^x + 2e^x and then I apply the formula for finding a equation of a line.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by goldenroll View Post
    Oh so on the first one am i using the chain rule? since its 2xe^x + 2e^x and then I apply the formula for finding a equation of a line.
    No, not quite. You use the product rule. IE: \frac{d}{dx}[u\cdot{v}]=u\cdot{v'}+v\cdot{u'}

    In this case, u=2x, and v=e^x
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