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Math Help - Need help please ! [urgent]

  1. #1
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    Exclamation Need help please ! [urgent]

    Hi there

    Please help me solve these problems. I tried, but I can't figure them out.

    1) If y is a differentiable function of x then the slope of the tangent to the curve  xy-2y+4y^2=6 at the point where [tex]y=1[tex] is..?


    2)Find the point on the graph of y=x^(1/2) between (1,1) and (9,3) at whtich the tangent to the graph has the same slop as the line through (1,1) and (9,3)

    3) Let f and g be differential functions such that :
    f(1) = 4
    g(1) = 3
    f'(3) = -5
    f'(1)= -4
    g'(1)= -3
    g'(3) = 2

    If h(x) = f(g(x)), then h'(1) =

    4) Find the derivative and the equation of the tangent line to  3x^2-3xy+2y^2=2

    5) Find the second derivative of y^2=x^2-2x

    6) Use the tangent line approximation of y= square root of x at x=16 to find the approximate value of the square root of 17

    And also, what does differentiable mean? How can you tell on a graph if a point is differentiable or not?

    Thanks for the help
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  2. #2
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    Quote Originally Posted by NoiseandAttack View Post
    Hi there

    Please help me solve these problems. I tried, but I can't figure them out.

    1) If y is a differentiable function of x then the slope of the tangent to the curve  xy-2y+4y^2=6 at the point where [tex]y=1[tex] is..?
    first find the corresponding x-value when y = 1 from the given equation.

    then differentiate implicitly and solve for y'. then plug in y = 1 and the corresponding x-value to solve

    can you continue?
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    Quote Originally Posted by Jhevon View Post
    first find the corresponding x-value when y = 1 from the given equation.

    then differentiate implicitly and solve for y'. then plug in y = 1 and the corresponding x-value to solve

    can you continue?
    What I have a problem with is finding the derivative of the equation. I'm not really sure how. Can you explain please?
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  4. #4
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    Quote Originally Posted by NoiseandAttack View Post
    2)Find the point on the graph of y=x^(1/2) between (1,1) and (9,3) at whtich the tangent to the graph has the same slop as the line through (1,1) and (9,3)
    find the slope of the line that passes through (1,1) and (9,3) (do you know how to do that?)

    then, find the derivative of y = x^(1/2) and set it equal to the value you found above. solve for x, this will give you the x-value of the point you are after. then plug this x-value into the original function to solve for the corresponding y-value, and then you'll have your point
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    Quote Originally Posted by NoiseandAttack View Post
    What I have a problem with is finding the derivative of the equation. I'm not really sure how. Can you explain please?
    remember, when doing implicit differentiation, we keep track of what we are differentiating and what we are differentiating with respect to. here we are differentiating with repsect to x. so when we differentiate an x-term, we attach dx/dx to it (since we differentiated an x-term with respect to x). but derivative notations can function as fractions and this cancels and we get 1, so we don't write it. when we differentiate a y-term, we attach dy/dx (since we differentiated a y-term with respect to x). we justify doing this by the chain rule. you can look it up in your text for more info.

    i will use y' to mean dy/dx to save on typing

    xy - 2y + 4y^2 = 6

    differentiating implicitly we get:

    y + x~y' - 2~y' + 8y~y' = 0 ...........note that i used the chain rule for the first term

    now, solve for y' and plug in the required x and y-values
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    Quote Originally Posted by NoiseandAttack View Post
    3) Let f and g be differential functions such that :
    f(1) = 4
    g(1) = 3
    f'(3) = -5
    f'(1)= -4
    g'(1)= -3
    g'(3) = 2

    If h(x) = f(g(x)), then h'(1) =
    Let h(x) = f(g(x))

    by the chain rule:

    h'(x) = f'(g(x)) \cdot g'(x)

    now, plug in x = 1 and solve
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    Quote Originally Posted by NoiseandAttack View Post
    4) Find the derivative and the equation of the tangent line to  3x^2-3xy+2y^2=2
    finding the derivative is very similar to the first question.

    to find the tangent line, we need to know at what point. when you have the point, use the point-slope form:

    y - y_1 = m(x - x_1)

    where m is the slope of the tangent line at the point (x_1,y_1) and (x_1,y_1) is a point the line passes through

    just solve for y
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    Quote Originally Posted by NoiseandAttack View Post
    5) Find the second derivative of y^2=x^2-2x
    we can use implicit differentiation here, but i think it will be too much work

    rewrite as y = \pm \sqrt{x^2 - 2x}

    and differentiate twice using the chain rule. you will also need the product rule for the second derivative
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    Quote Originally Posted by NoiseandAttack View Post
    6) Use the tangent line approximation of y= square root of x at x=16 to find the approximate value of the square root of 17
    this is the formula for the linear approximation of a function.

    f(x) \approx f(a) + f'(a)(x - a)

    here x is the value for which we want to find f(x) for, but we can't because it does not have a nice answer, and a is a point close to x for which we do know the exact value of the function.

    here f(x) = \sqrt{x}, x = 17 and a = 16

    thus, \sqrt{17} = f(17) \approx f(16) + f'(16)(17 - 16)

    now continue
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    Quote Originally Posted by NoiseandAttack View Post
    And also, what does differentiable mean?
    "differentiable" means we can differentiate it. that is, the function's derivative can be found, or exists at a certain point, or on some interval.

    How can you tell on a graph if a point is differentiable or not?
    recall how you defined the derivative. we started out with the slope formula for the equation of a line. thus, in layman's terms, a function is differentiable at a point, if it is approximately linear at that point. that is, if we zoom in close to a point, the function looks like a straight line. what this translates to is the following.

    - we cannot find the derivative at points where a function takes a sharp turn, e.q. y = |x| at x = 0

    - we cannot find the derivative of a function where it is discontinuous
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  11. #11
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    Can you show the steps for finding the implicit derivative, please?

    Thank you for all the other help, though.
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  12. #12
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    Quote Originally Posted by NoiseandAttack View Post
    Can you show the steps for finding the implicit derivative, please?

    Thank you for all the other help, though.
    what are you talking about? in post #5 i did show all the steps. all that was left for you to do is solve the equation for y'
    Last edited by Jhevon; November 25th 2007 at 07:09 PM.
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  13. #13
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    Oohh, I'm so sorry.. I missed that !

    Thanks for the help !
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  14. #14
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    Quote Originally Posted by NoiseandAttack View Post
    Oohh, I'm so sorry.. I missed that !

    Thanks for the help !
    ok. you're welcome

    if you have any problems, say so
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