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calculus

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Find the equation of the tangent line to the curve y=sqrt(x) at x = c. Put the answer in slope intercept form y=mx+b. Use the definition of the slope of the tangent with no short cuts.

  • calculus - ,

    so the point of contact is (c, √c)

    y = √x = x^(1/2)
    dy/dx = (1/2)x^(-1/2)
    at our point
    dy/dx = (1/2)c^(-1/2) = 1/(2√c) = slope = m

    y = (1/(2√c) x + b
    at(c, √c)
    √c = (1/(2√c)(c) + b
    times 2√c
    2c = c + 2√c b
    c = 2√c b
    b = c/√c

    y = (1/(2√c)) x + c/√c

    check my arithmetic
    You could rationalize the denominator if you had to.

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