1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook