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Homework Help: Intro to Derivatives (Help)

  1. Oct 23, 2008 #1


    Haha I tried to make it as neat as possible... Please help me , this isn't homework, but I just want to learn the steps so I'll be ready for my test in two weeks. I posted in this forum because when I posted it on the analysis part, they moved it to the Homework section for some reason. The blue is the ones I'm confident about, and the Red is the ones I'm having trouble with and need help explaining.

    I know this is a lot, so I really appreciate it if someone is willing to help

    thanks in advance :)
  2. jcsd
  3. Oct 23, 2008 #2


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    Hi asdfsystema! :smile:

    1. -9 is right.

    But how did you get m = 3? :confused:
  4. Oct 26, 2008 #3
    I'm not sure what i'm supposed to do ... did you read the part in red ? I dont know if I'm supposed to find the tangent line to the equation of the derivative or the one in the question...
  5. Oct 26, 2008 #4


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    Your supposed to find the tangent line to [itex]y=4-3x^3[/itex].
  6. Oct 26, 2008 #5


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    Remember, the tangent line to a function y(x) at the point [itex](x_0,y_0)[/itex] is parallel to [itex]y'(x_0)[/itex]....does this help you?
  7. Oct 26, 2008 #6
    ok so for #1

    y= 4-3x^3 (1,1) I found the derative y'=-9x and then plugged in 1 to get the slope m= -9.
    then using point slope formula, (y-1)= -9(x-1) and got y= -9x + 10. so m= -9 and b= +10

    derivative of y=5/x-2 is y'= -5/x^2-4x+4. and i plugged in 5 to get slope m= -5/9. I used the point slope formula (y-1.66)= -5/9 (x-5) and got y= -5/9x+25/9+1.66 (can anyone tell me what the fraction is for 25/9+1.66666?

    Also help me out with #3-8 please =)
  8. Oct 26, 2008 #7


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    #1 looks correct

    for #2, use the fact that 1.6666...=5/3=15/9

    for #3-8....what is the slope of a horizontal line? what does that mean the slope of the tangent at (x_0,y_0) is if the tangent there is horizontal?
  9. Oct 26, 2008 #8
    the slope of a horizontal liine is 0 so the slope of the tangent there is 0 ?

    but what part of the equation gives me clues where the negative / positive horizontal line is ?

    wait.. is the answer 0 for both ?
  10. Oct 26, 2008 #9


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    The slope of any horizontal tangent is zero. So, if the horizontal tangents are located at the points (x_0,y_0) and (x_1,y_1) then y'(x_0)=0 and y'(x_1)=0.

    For #3, y(x)=2x^3 +12x^2-72+8. What is y'(x) then? Where does y'(x)=0?
  11. Oct 27, 2008 #10


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    No, since that says a line is parallel to a number, I don't think it will help anyone.
  12. Oct 27, 2008 #11
    i am getting very confused , hallsofivy , what is it that im supposed to do ?
  13. Oct 27, 2008 #12


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    Hi asdfsystema! :smile:

    He means that the slope (the tan) of the tangent line is equal to y'(x0) :smile:
  14. Oct 27, 2008 #13

    for number #3, what i did was find the derivative of the equation f(x)= 2x^3+12x^2-72x+8 which is f'(x)= 6x^2+24x-72 and I divided that by 6 and got x^2+4x-12 and I factored them both and got (x+6)(x-2) , does that mean the horizontal tangents are at x= -6 and x= +2 or do I need to do one more step?

    #4 I thought about it and first step I did was since f(x)= x^9, I took the derivative of that which gave me f'(x)=9x^8. then I plugged it into f'(-1)= 9(-1)^8 = 9. and since h'(-1)= 5 , I just took that and multiplied 9x5 = 45..

    Please check my answers for 5-8 , I think i did it correctly.

    for #9 , I figured out the derivatives , but none of them match. I think I could be doing something wrong ?

    #10-11 , i'll appreciate it if someone can verify the answers and if possible list a quicker step into solving it ..

    #12 and #13 im completely clueless T__T ..

    thank you so much !
  15. Oct 28, 2008 #14
    anymore help please ? i have a quiz this coming wednesday on this material ):
  16. Oct 28, 2008 #15


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    That's fine. :smile:
    Nooo … :frown: use the product rule.
    Yes … you need to learn your trigonometric identities …

    #9 isn't really a calculus problem, its a trig problem …

    for example, you should automatically know sec2x = 1 + tan2x.
    Sorry, your #10 is rubbish … you're obviously confused about the chain rule … the derivative of sinx is cosx.

    For #11, use the product rule … (fg)' = f'g + g'f.
    #12 … the slope of the tangent of f(x) is the derivative, f'(x).

    #13 … f(g(x))' = f'(g(x)) times g'(x) … in this case, g'(x) = 4.

    Your H'' is fine. :smile:
  17. Oct 28, 2008 #16
    thank you so much. ! one last thing, can you double check my answer sheet?

    1. h'(1)= -9 m=-9 b=10
    2. f'(5) = -5/9 m=-5/9 b=40/9 after using point slope
    3. x=-6 x=2

    4. i used your advice but not sure if i did it incorrectly..
    f(x)=x^9(h(x) so u= x^9 u'=9x^8 v= h(x) and h(-1) = 2 v'= h'(x) and h'(-1) =5 so I leave that
    I simply plugged in the product rule 9(-1)^8 * 2 + (-1)^9 * 5 and got 13 as answer

    5. f(x)= (7x^2-7) (3x+2) find f'(x) . f'(x) = (14x)(3x+2) + (7x^2-7) (3) = 63x^2 + 28x-21
    f'(2) =287 ???

    6. f(x)= 5x+5/7x+5 find f'(x) = (5)(7x+5)-(5x+5)(7) / (7x+5)^2 ===>
    -10/49x^2+70x+25 f'(2) = -10/361???

    7. f(t)= (t^2+7t+7) (2t^2+6)
    f'(t)= 8t^3+42t^2+40t+42
    f'(3) = 756 ??

    8.quotient rule f(x)=5x^2+3x+3/ sqrt(x)
    f'(x)= (10x+3)(x^1/2)- (5x^2+3x+3) (-x^-1/2) / x
    f'(16)= 102.29 Is this considered 4-5 sig figs ? Is there another way to do this problem that will result in a fractional answer?

    9. 1. C 2. D 3.A 4.B ??

    10. f(x)=7sinx+7cosx find f'(x)
    f'(x)= 7cosx-7sinx . agh this was an easy problem too
    f'(pi/3)= 7(.5)-7(.866)= -2.5621 (is this 5 significant figures?)

    11. f(x)= -3x(sinx+cosx) product rule
    u=-3x u'=-3 v= sinx+cosx v'=cosx-sinx
    f'(x)= (-3sinx-3cosx)+(-3x)(cosx-sinx)... is this the same as (3x-3)sinx+(-3x-3)cosx??
    f'(pi/3)= -5.2479 ?
    wow is there anything I can do with these decimals lol

    12.y=6secx-12cosx at points (pi/3,6) find y' and then tangent line
    y'= 6secxtanx+12sinx
    y'(pi/3)= 31.1769
    (y-y1)= m(x-x1)
    (y-6)=31.1769(x-pi/3) ---> y=31.1769x-26.6483
    m= 31.1769x b= -26.6483

    13. h(t)=tan(4t5) h'(t) use chainrule
    h'(t)= sec^2(u) * du/dx = 4sec^2(4t+5) h'(4) = 10.4256 ??? are these 4-5 sig figs ?

    h''(t) use chain rule again
    h''(t)= 4(2*sec^2(u)*tan(u)) * du/dx ==> 8sec^2(4t+5)*tan(4t+5)*4 ==>
    32sec^2(4t+5)*tan(4t+5) . h''(4) = -162.9185??? are these 4-5 sig figs

    seriously ... my teacher is crazy, I cant find any fraction for these things and there are a lot of decimals. Is it possible to condense? If not, then are these 4-5 sig figs? thanks a lot tim !! you are the best ^^
  18. Oct 28, 2008 #17
    please check my answers ! Test tomorrow :(
  19. Oct 29, 2008 #18
    Guys, please just help me double check :)
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