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Math Help - Tangent line to a curve.

  1. #1
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    Tangent line to a curve.

    I'm trying to find the equation of the line tangent to the curve of y=ln(sqrt(7x^2+9)) at x=1.

    So I need to find the first derivative:

    y'= 7x/(7x^2+9) and at x=1 I find y= 16 (1,16)

    Because y= (approx.) 2.77 at x=1 I have a point to substitute in my equation of the line:

    y=16x+b

    and I find b= -16.23 (approx.)

    then the equation of my tangent line at 1 is y=16x-13.23

    except that looks nothing like the tangent line on my graph. Where am I off (way off??)
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Tangent line to a curve.

    Your derivative is fine.

    y'(1) = \dfrac{7}{16} - this is your gradient at x=1 (in other words the slope of the tangent line at x is y'(x))

    The point on f(x) is f(1) = \ln(\sqrt{7+9}) = \ln(4) = 2\ln(2) giving the point (1, 2\ln(2)) to use below


    You can then use the equation of a straight line with what you found above.
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