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A complicated derivative using chain rule

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a function z, and I need to find the derivative dz/dt "using the chain rule without substitution"

    2. Relevant equations

    [tex]z = x^{2}y^{3} + e^{y}\cos x[/tex]
    [tex]x = \log(t^{2})[/tex]
    [tex]y = \sin(4t)[/tex]

    3. The attempt at a solution

    [tex]\frac{\mathrm{d} z}{\mathrm{d} t} = 2xy^{3}+3x^{2}y^{2}+e^{y}\cos x-e^{y}\sin x[/tex]
    [tex]\frac{\mathrm{d} z}{\mathrm{d} t} = 2\log(t^{2})\sin(4t)^{3}+3(\log(t^{2}))^{2}(\sin(4t))^{2}+e^{\sin(4t)}(\cos (\log(t^{2}))-\sin (\log(t^{2})))[/tex]
    [tex]\frac{\mathrm{d} z}{\mathrm{d} t} = 4\log(t)\sin(4t)^{3}+3(2\log(t))^{2}(sin(4t))^{2}+e^{sin(4t)}(\cos (2\log(t))-sin (2\log(t)))[/tex]

    I'm a bit stuck at this point, I feel like I should be able to do something with the log^2 and the e^sin . (cos(log) - sin(log))?
     
    Last edited: Mar 26, 2012
  2. jcsd
  3. Mar 27, 2012 #2
    That's not right.. Now you have x=x(t), y=y(t), z=z(t) so you'd get
    [tex] \frac{dz}{dt} = 3 x^2 y^2 \frac{dy}{dt} + ... [/tex]
    etc. This is the chain rule, and it's the most important differentiation rule, so you should definitely read on it further :-)
     
  4. Mar 29, 2012 #3
    I just wanted to say thanks, I studied my book, and I did approach this in the wrong way! I won't do the latex script, but my answer is (without substitution):

    dz/dt = ∂z/∂x . dx/dt + ∂z/∂y . dy/dt

    = 2e^xy .(2x + x^2 .y)(t) - (x^3 .e^xy)(-t^-2)
     
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