A complicated derivative using chain rule

[Adyssa]

Homework Statement

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

Homework Equations

$$z = x^{2}y^{3} + e^{y}\cos x$$
$$x = \log(t^{2})$$
$$y = \sin(4t)$$

The Attempt at a Solution

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

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))?

 

[clamtrox]

That's not right.. Now you have x=x(t), y=y(t), z=z(t) so you'd get
$$\frac{dz}{dt} = 3 x^2 y^2 \frac{dy}{dt} + ...$$
etc. This is the chain rule, and it's the most important differentiation rule, so you should definitely read on it further :-)

 

[Adyssa]

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)