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Differentiation - finding equation of normal to curve

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the normal to the curve y=(6x+3)^(1/2) at the point for which x=13.

    2. Relevant equations



    3. The attempt at a solution


    gradient of normal = -3(6x+3)^2

    at x=13, dy/dx=-3(6x13+3)^2=19683.

    obviously a silly answer and the textbook disagrees.

    So at this point I could simplify y=(6x+3)^(1/2) to y=(2x+1)^2 maybe? Even after doing this, I end up with an even slightly smaller silly answer and the textbook still disagrees, unless the textbook is wrong, although this is highly unlikely.

  2. jcsd
  3. Oct 13, 2011 #2


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    You will want to start by checking this:
  4. Oct 13, 2011 #3
    I inputted y=(6x+3)^(1/2) into wolfram alpha and the answer came out to be:


    I'm confused. I used the chain rule to get:

    subu=6x+3 so that y=u^1/2
    dy/du=1/2u^-(1/2)=1/2(6x+3)^-(1/2) and du/dx=6

    => dy/dx=dy/du*du/dx=3(6x+3)^-(1/2).

    Last edited: Oct 13, 2011
  5. Oct 13, 2011 #4
    Work out the derivative by hand. It's straightforward.
  6. Oct 13, 2011 #5


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    I put that into WolframAlpha and got the correct answer.
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