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Math Help - Derivs heeelp

  1. #1
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    Derivs heeelp

    f(x)=[(x+2)/(x+1)](2x-7)

    I keep getting [-2x^2+7x-3] / (x+1)
    but that is not one of the answer choices.

    Also, (x-2)/sqrt(x^2-1)

    Find an equation for the tan line to the graph f at the point P(pi/4, f(pi/4)) when f(x)=9tan x

    Let f(x) = (4x-1)^6(3x+3)^2 Find the x-intercept of the tan line to the graph of f at the pt where the graph of f crosses the y-axis
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  2. #2
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    Hi

    Quote Originally Posted by bballa273 View Post
    f(x)=[(x+2)/(x+1)](2x-7)

    I keep getting [-2x^2+7x-3] / (x+1)
    but that is not one of the answer choices.
    this is indeed really hard

    First: Quotient rule

     [(x+2)/(x+1)]' = \frac{1*(x+1)-1*(x+2)}{(x+1)^2}=:v'

    ----
    (2x-7)' = 2 =: u'

    Product rule (u*v)' = u'*v + v'*u

    and so on...

    But it is much easier, if you multiply (x+2)(2x-7) first. Then you only must use the quotient rule


    Quote Originally Posted by bballa273 View Post
    Also, (x-2)/sqrt(x^2-1)
    quotient rule:   (\frac{u}{v})' = \frac{u'*v-v'*u}{v^2}

    here: u= x-2 => u'=1

    v = \sqrt{x^2-1} = (x^2-1)^{0.5}

    v' = 0.5 * 2x * (x^2-1)^{-0.5}

    ...

    Quote Originally Posted by bballa273 View Post
    Find an equation for the tan line to the graph f at the point P(pi/4, f(pi/4)) when f(x)=9tan x

    Let f(x) = (4x-1)^6(3x+3)^2 Find the x-intercept of the tan line to the graph of f at the pt where the graph of f crosses the y-axis
    I dont know. Maybe someone else does

    Rapha
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  3. #3
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    Quote Originally Posted by bballa273 View Post
    [snip]
    Find an equation for the tan line to the graph f at the point P(pi/4, f(pi/4)) when f(x)=9tan x

    [snip]
    You know two things about the line:

    1. Point on line is \left(\frac{\pi}{4}, \, 9 \right) (why?)

    2. Gradient of line is 9 \sec^2 \left( \frac{\pi}{4} \right) = \, .... (why?)

    So you should be able to get the equation of the line.
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  4. #4
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    Quote Originally Posted by bballa273 View Post
    [snip]
    Let f(x) = (4x-1)^6(3x+3)^2 Find the x-intercept of the tan line to the graph of f at the pt where the graph of f crosses the y-axis
    1. Find the y-intercept of y = f(x). This gives you a known point on the tangent line.

    2. Evaluate the derivative of f(x) at the y-intercept. This gives you the gradient of the tangent line.

    3. Use 1. and 2. to get the equation of the tangent line.

    4. Use the equation in 3. to get the x-intercept of the tangent.
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