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Homework Help: Easy First year Calc - I need someone to mark

  1. Apr 23, 2009 #1
    i don't really have time to write out all my awnsers, but would appreciate if someone could give me the awnsers to these questions

    i'm confident in my maths but just need someone to give me a rundown of the awnsers, because the book i'm using dosen't have any!

    there are 12 questions in all, and most are REALLY easy and i really want to know if i got them right or not

    1: f(x) = x sin x

    2: y = (1 + sin x) / (x + cos x)

    find an equation of the tangent line to the curve at the given point
    3: y = (1+ x) cos x, (0,1)

    identify the inner function, outer function and find the derivative dy/dx
    4: Squareroot(4+3x)

    5: f(x) = (x^2 - x +1)^3

    6: y=(a^3) + ((cos^3)*x)

    find an equation of the tangent line to the curve at the given point
    7: y = squareroot(5+x^2), (2,3)

    find y by implicit differentiation
    8: (4x^2 )+ (9y^2) = 36
    then solve the equation explicityly for y and differentiate to get y' in terms of x

    then check your solutions for 8:/9: are consisteny by substituting the expression for y into your solution for 8:
    (n.b this is the one i had trouble with, 8,9,10)

    find dy/dx by implicit differentiation
    11: x^2 - 2xy + y^4 = c
    (took me 10 minutes lol)

    use implicit differentiation to find an equation tangent to the line of the curve to the point
    12: x^2 + 2xy - y^2 + x =2
    Last edited by a moderator: Apr 23, 2009
  2. jcsd
  3. Apr 23, 2009 #2
    oh and thanks XD (sorry about that TEX thing it's still not workin)
  4. Apr 23, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's not the way it works here. You post your answers and we'll tell you if they're correct.

    If they're not correct, we might also be nice enough to help you get the right answers.

    P.S. In future, please don't use large text, it makes your post difficult to read.
  5. Apr 23, 2009 #4
    yeah sorry about that, as you can see from the post it was almost 4am and i was getting kind of tired so i didn't want to write all the awnsers up,

    i'll make sure i won't do that again though, thanks XD anyways

    my attempts:
    differentiate xsinx = xcosx

    differentiate sinx + 10tanx = cosx + 10*sec(x)^2

    dy/dx of (1+sinx)/(x+cosx) =
    (x+cosx)*(1+sinx)' - ( 1+sinx) * (x+cosx) /(x+cosx)^2
    = (x+cosx)(cosx) - (1+sinx)(1-sinx) / (x+cosx)^2

    find an equation of the line to tangent given coords
    y=(1+x)(cosx) , (0,1)
    y = mx+c , 1 = 1*cos0)(-sin0) = 1? not completely sure on this one

    differentiate y= squareroot of (4+3x)
    (1/2 * (4+3x) ^ -1/2 )*(3)

    sorry i'll write up the other equations for you later, got a lecture just now
  6. Apr 23, 2009 #5


    Staff: Mentor

    Two problems above. First, xcosx is not the derivative of xsinx. Use the product rule.
    Second, notation. An improvement would be to write d/dx(xsinx) = <the right answer>

    The answer is correct, but you ought to say this:
    d/dx(sinx + 10tanx) = cosx + 10*sec(x)^2
    Then it's clear that you have taken the derivative of what's on the left side to get what's on the right side.
    You have some mistakes in the above, but you're on the right track with the quotient rule. Your formatting leaves something to desire, since the division is shown as occurring for the product to the right of the minus sign, and that's not what you meant, but that's the way most of us here would interpret it. To fix that, you need one more pair of parentheses around the two products in the numerator. Like this, in other words:
    [(x+cosx)(cosx) - (1+sinx)(1-sinx)] / (x+cosx)^2

    As already noted, this isn't the right answer, but you should be able to fix it without too much work.

    Also, you don't take "dy/dx" of something. dy/dx is already the derivative of y with respect to x. The operator that indicates you're going to take the derivative is d/dx.

    dy/dx = d/dx((1+x)(cosx))
    Use the product rule on the right side, then evaluate the derivative (dy/dx) at x = 0.
    Looks good, but you want to say dy/dx = (1/2 * (4+3x) ^ (-1/2) )*(3), and this can be simplified considerably as
    dy/dx = 3/2sqrt(4 + 3x).
    I tried to write this in LaTeX, but it still appears to not be working.
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