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Finding the Derivative

  1. Dec 5, 2005 #1
    Hey.. I'd appreciate it if you could guys could verify my answers for the following questions..

    Find dy/dx:

    1) y=arcsin(2x)
    y'= (2)/sqrt(1-4x^2)

    2) y=arctan(x^2)
    y'= 2x/(1+x^4)

    3) 3sqrt(x)=sqrt(y)
    dy/dx=( 3.(x)^(-.5) / (y)^(-.5))

    4) xy+y=x
    dy/dx = (-y)/(x+y)

    5) e^cosy=x^3 arctan y

    (e^cosy) (-siny. dy/dx ) = 3x^2 + (1/1+y^2)dy/dx

    6) Find slope of the following when x equals the indicated value.

    A) xy+x+y=8, x=2

    dy/dx= (-y-1)/(x+Y)
    dy/dx= -3/4

    B) sin^2y + cos^2x=4, x=(pi/2)

    9) 2x^2+y^2=4
    Dy/dx= -2x/y

    10) ln(xy)=4
    ln ( x. dy/dx + y ) = 0
    1/ x.dy/dx + y = 0
    dunno how to proceed
    dunno how to do this..

    thanks much
    Last edited: Dec 5, 2005
  2. jcsd
  3. Dec 5, 2005 #2
    When writing your answer, always try to write y in terms of x. However, if y is too big to write then don't care.

    for number 3 do you mean
    [tex] \sqrt{x}[3] = \sqrt{y} [/tex] OR
    [tex] 3 \sqrt{x} = \sqrt{y} [/tex]
    in either case you are wrong.
    if it was the first one then i suggest you SQUARE both sides and differentiate
    in teh second one i suggest you do the same.

    for number 4
    xy + y = x , isolate for y before you differentiate. You can actually do that for this problem. SOmetimes you cannot in which case you have to do it implicitly

    for number 10
    How do you differentiate a logarithm?
    THis is how i remember - this is not to be taken as theoretical or anything technical
    (FLip the argument, apply natural log to the base of the logarithm and place taht in teh denominator, then differentiate the argument of the logarithm itself.)

    Argument = Stuff to which the logarithm is being applied.

    Hope this is useful
  4. Dec 7, 2005 #3
    hey thanks for your help..

    I needed help with some other probs .. i'd appreciate your help

    Water runs into a conical tank at the rate of 2 cubic feet per minute. The tank stands point down and has a height of 10 feet and a maximum radius of 5 feet. How fast is the water level rising when the water is 6feet deep?



    B=Pi r^2
    volume of cone = 1/3 B.H
    V=1/3 (pi.r^2)(2r)
    v= 2/3 ( pi. r^3)

    now I need to find dv/dt when H=6, r = 3

    dv/dt = 1/12 pi h^2
    dv/dt = 0.03 feet / min approx

    is trhat correcT?


    1) x+sin(y)= Pi/2
    1+cosy dy/dx = 0
    dy/dx = arccosy -1

    2) tan (x+y)=x
    x+y = arctan x
    1+dy/dx = 1/(1+x^2) . 2x
    dy/dx = 2x/(1+x^2) -1

    3) cos(x)+cos(y)=2y
    -sinx -siny dy/dx = 2
    dy/dx = arcsiny (-2-sinx)

    4) xy=e^(xy)
    dy/dx = (e^(xy) - y )/( x)

    5) taken from my first set of questions:

    e^cosy= x^3 arctany
    e^(Cosy) . ( -siny) = 3x^2 arctany + 1/1+y^2.dy/dx . x^3
    therefore, dy/dx = (e^cosy. (-siny) - 3x^2arctan y . ( 1+y^2) ) / (x^3)
    Last edited: Dec 7, 2005
  5. Dec 10, 2005 #4
    for the related rates problem (the water draining in the tank) there is a mistake: when you take the derivative in respect to time, you forgot to include dh/dt. In addition, you already know dV/dt, so you don't solve for it. The problem includes that the rate that the volume changes is 2 ft^3/min.

    it should read:
    [tex] V= \frac {1}{12} \pi h^3 [/tex]
    [tex] \frac {dV}{dT} = \frac {\pi h^2}{4} \frac {dh}{dt} [/tex]
    Last edited: Dec 10, 2005
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