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Find the equation of tangent line for derivatives of functions

  1. May 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the tangent line at the point indicated

    f(x) = 5x2-2x+9 , x = 1

    2. Relevant equations

    (d/dx) bx = ln(b)bx

    General Power Rule which states:

    (d/dx) g(x)n = n(g(x))n-1 * g'(x)

    3. The attempt at a solution

    So looking at a previous problem first, we have
    s(t) = 37t, t = 2

    So, we can substitute u for the exponent, and take the derivative of that. We get
    (ln 3) * 3u * (7t)' which equals (7 ln 3)314

    Since it wants to find the tangent line we have to take it further. The equation to the tangent line is y-f(a) = f'(a)(x-a) and point slope form is y = f'(a)(x-a) + f(a)

    It helps me to look at it in this scenario as f'(a) = solution before putting it in slope form, a = the given x value(or t in this case), and f(a) = bx

    Using this first example problem that would mean

    y = 7 ln 3 * 314(t-2) + 314, and the book confirms this.

    Now, trying to generalize these principles to the problem I am working on..

    I switch the exponent with "u" to simplify the problem. We now have (ln 5)5u(x2-2x+9)' and since x = 1 its

    (ln 5)5u * 8 which gives us (8 ln 5)58

    Attemping to put this into the equation for a tangent line,

    f'(a) = (8 ln 5)58
    a = 1
    f(a) = bx = 58

    so putting that all together is

    y = (8 ln 5)58(x-1) + 58....

    or so I thought. The book gives

    y = 58

    How did I screw this up?
  2. jcsd
  3. May 15, 2013 #2
    Ok, I think I know at least the beginning of where I went wrong.

    We now have (ln 5)5u(x2-2x+9)' and since x = 1 its (ln 5)5u * 8 which gives us (8 ln 5)58

    The derivative of (x2-2x+9) isn't 8. It's 2x-2. I think this is where I need to start.
  4. May 15, 2013 #3


    Staff: Mentor

    f'(1) = 0, which makes the tangent line horizontal.

    If f(x) = 5x2- 2x + 9 , then f'(x) = (2x - 2) 5x2- 2x + 9, so f'(1) = 0.
  5. May 15, 2013 #4


    Staff: Mentor

    It's a lot simpler to convert exponential functions to one that involves e.

    For example, 5 = eln(5), so 5x = (eln(5))x = ex ln(5).

    Differentiation is pretty easy, using the fact that d/dx(eu) = eu * du/dx.
  6. May 15, 2013 #5


    Staff: Mentor

    Yes. You are combining formulas for f'(x) and f'(a) for some number a. Keep them separate, and replace x by a only at the end.
  7. May 15, 2013 #6
    Yes, it makes complete sense now. It ends up

    (2x-2)(ln 5)58

    I believe you can write that out as

    2x ln 5 * 2 ln 5 and, x is 1, so

    2 ln 5 - 2 ln 5 = 0

    Zero times 58 is zero. Got it.
  8. May 15, 2013 #7


    Staff: Mentor

    There's no need to expand the expression. Since x = 1, we can see immediately that 2x -2 = 0.

    BTW, in the above, "right" is wrong but "write" is right.
  9. May 15, 2013 #8
    Yeah, I caught it just before you posted that lol. Thanks Mark.
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