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Homework Help: Basic Calc prob :-s

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data

    for each of the following find dy/dx, expressing the answer as a function of x.

    (A) y=z7, z=sin(x)

    (B) y=√z , z=x4+1

    2. Relevant equations



    3. The attempt at a solution

    I'm having some trouble working out what the question is actually asking me to do... If i go at it normally i get for eg 7z6cos(x) ... but this isnt the answer.
     
  2. jcsd
  3. Apr 13, 2010 #2

    Doc Al

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    Staff: Mentor

    Express the answer as a function of x. (Get rid of that z!)
     
  4. Apr 13, 2010 #3
    so i'm literally just replacing the z?

    sin(x)7sin(x)

    becomes

    7sin(x)6cos(x)
     
  5. Apr 13, 2010 #4

    lanedance

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    Homework Helper

    the 2nd line looks ok, but why do you start with sin(x)7sin(x)?

    also to help people help you, try & be clear in what you want to communicate and try and give all the info eg.

    so substituting in for z(x) = sin(x) gives
    y(x) = sin(x)7
    then differentiating w.r.t. x
    y'(x) = 7sin(x)6cos(x)
     
  6. Apr 13, 2010 #5
    i was replacing the z... prob mixed up the steps

    Are these questions based on the chain rule?
     
  7. Apr 13, 2010 #6

    Doc Al

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    Yes. Your answer of 7z6cos(x) becomes 7sin6(x)cos(x).
     
  8. Apr 13, 2010 #7

    Doc Al

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    Yes.
     
  9. Apr 13, 2010 #8

    lanedance

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    yes & and that's exactly what you've used
     
  10. Apr 13, 2010 #9
    ok cool, because i was doing some work on products and quotients and then got hit with these so was a little thrown.

    thanks for your help!

    ps Any help with (B) :p
     
  11. Apr 13, 2010 #10

    lanedance

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    in essence same as a) using chain rule
     
  12. Apr 13, 2010 #11

    lanedance

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    in both cases you are given y(z) and z(x), then substitute to get y(z(x))

    the derivative of y w.r.t. the chain rule is
    [tex] \frac{dy(z(x))}{dx} = \frac{dy(z)}{dz} \frac{dz(x)}{dx} [/tex]

    noting as you did early on you need to substitute in for z = z(x) in the final answer if any remain
     
  13. Apr 13, 2010 #12
    ok im kinda struggling with the latex code... what a ballache! apologies if this is confusing

    z=1/2√z & z=x4+1 = 4x3

    then i get

    4x3/2√x4+1

    how does that look...?
     
  14. Apr 13, 2010 #13

    Mark44

    Staff: Mentor

    It should be y = (1/2)sqrt(z) + 1. You don't want z to be defined as the square root of itself. Then, if z = z=x4+1, dz/dx = 4x3, not z.
    Other than it's incorrect? It should be written as an equation that identifies what you have found.

    You might find it easier to work with exponents rather than radicals. y = (1/2)z1/2. dy/dx = dy/dz * dz/dx
     
  15. Apr 16, 2010 #14
    ok. thanks for the help dude
     
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