Results 1 to 5 of 5

Math Help - Differential Calculus midterm

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    5

    Differential Calculus midterm

    Hey guys can you help me with the questions on this Math midterm?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    What exactly do you need help with? This is one of your first posts, and you simply post the question and ask for help. It seems like you just want the answers.

    For example, the question: Find the derivative of f(x)=\frac{x}{1+x} I would treat it as a product like f(x)=x*(x+1)^{-1}, and then you find the derivative by applying the product and power rules.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2008
    Posts
    5
    I don't want the answers, I'm just completely clueless as to how I would approach them.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by VX-1 View Post
    Hey guys can you help me with the questions on this Math midterm?

    1.
    "orthogonal" means "perpendicular." so you want all the points on
    y = 5x^3 + 5 where the slope (that is, the derivative) is 30. please tell me you know why we need it to be 30


    2.
    the definition says: f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h

    now plug in f(x + h) and f(x) and take the limit. (you will need to simplify before you can do that, of course)

    in case you have never seen that definition, you must have used this one. it is an alternate definition equivalent to the one above.

    f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}


    3.
    (a) note that \sqrt{x^2 + 1} increases a lot faster than \sqrt{x}. alternatively, note that as x \to \infty the +1 doesn't really matter. thus you have \sqrt{x^2} - \srqt{x} = |x| - \sqrt{x} = x - \sqrt{x} for positive x's

    (b) more straightforward solution: multiply by the cojugate of the denominator over itself.

    alternate solution: note that x - 4 = (\sqrt{x})^2 - 4 = (\sqrt{x} + 2)(\sqrt{x} - 2)


    4.
    you want to find a and b so that the function is continuous and the derivative limit for both parts exist and are equal.


    5.
    first find f'(x) (you will need the chain rule). then you can answer the problem. i hope you know how to tell something is an extreme value from the derivative, and that you know a function is increasing wherever the derivative is positive and decreasing where it is negative


    6.
    Let f(x) = \frac 1{x^2 + 1}

    by tweaking our definition for derivative a bit, we obtain the linear approximation formula:

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

    now, x is the value that we want to find f(x) for, while a is a value close to x that we already know f(a) and f'(a) for



    so there's how you should approach these problems. good luck!

    Quote Originally Posted by colby2152 View Post
    For example, the question: Find the derivative of f(x)=\frac{x}{1+x} I would treat it as a product like f(x)=x*(x+1)^{-1}, and then you find the derivative by applying the product and power rules.
    well, they said use the definition. i am pretty sure they are talking about using limits to calculate the derivative
    Follow Math Help Forum on Facebook and Google+

  5. #5
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    well, they said use the definition. i am pretty sure they are talking about using limits to calculate the derivative
    True. I missed that part of the question. You'll have to find the derivative via the tedious, tried and true method that is the definition of the derivative:

    f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. HELP in differential calculus
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 5th 2009, 04:32 AM
  2. Midterm...Calculus optimization problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2008, 06:04 AM
  3. Differential calculus?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 30th 2008, 09:58 AM
  4. Differential Calculus
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 2nd 2007, 08:11 AM
  5. Pre-Calculus Midterm
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: January 2nd 2007, 03:23 PM

Search Tags


/mathhelpforum @mathhelpforum