Answer:
a) [tex]\frac{dz}{dt} = e^{t}\cdot (2\cdot t+2\cdot t^{2}-1)[/tex]
b) [tex]\frac{dz}{dt} = [4\cdot \ln t\cdot \sin^{3} 4t - \sin 4t \cdot \sin (2\ln t)]\cdot \left(\frac{2}{t} \right)+[12\cdot (\ln t)^{2}\cdot \sin^{2} 4t + \cos (2\cdot \ln t)]\cdot (4\cdot \cos 4t)[/tex]
Step-by-step explanation:
In this exercise we use implicit differentiation to define [tex]\frac{dz}{dt}[/tex]:
a) [tex]z = x\cdot e^{x\cdot y}[/tex], [tex]x = t^{2}[/tex], [tex]y = t^{-1}[/tex]
[tex]\frac{dz}{dt} = \frac{dx}{dt}\cdot e^{x\cdot y} +x\cdot e^{x\cdot y}\cdot \left[\frac{dx}{dt}\cdot y+x\cdot \frac{dy}{dt} \right][/tex]
[tex]\frac{dz}{dt} = e^{x\cdot y}\cdot \left\{\frac{dx}{dt}+x\cdot \left[\frac{dx}{dt}\cdot y+x\cdot \frac{dy}{dt} \right] \right\}[/tex]
[tex]\frac{dz}{dt} = e^{x\cdot y}\cdot \left[(1+x\cdot y)\cdot \frac{dx}{dt}+x\cdot \frac{dy}{dt} \right][/tex](1)
[tex]\frac{dx}{dt} = 2\cdot t[/tex], [tex]\frac{dy}{dt} = -t^{-2}[/tex]
[tex]\frac{dz}{dt} = e^{t}\cdot \left[(1+t)\cdot 2\cdot t +t^{2}\cdot (-t^{-2})\right][/tex]
[tex]\frac{dz}{dt} = e^{t}\cdot (2\cdot t+2\cdot t^{2}-1)[/tex]
b) [tex]z = x^{2}\cdot y^{3}+y\cdot \cos x[/tex], [tex]x = \ln t^{2}[/tex], [tex]y = \sin (4\cdot t)[/tex]
[tex]\frac{dz}{dt} = 2\cdot x\cdot y^{3}\cdot \frac{dx}{dt} +x^{2}\cdot (3\cdot y^{2})\cdot \frac{dy}{dt}+(\cos x)\cdot \frac{dy}{dt} +y\cdot (-\sin x)\cdot \frac{dx}{dt}[/tex]
[tex]\frac{dz}{dt} = (2\cdot x\cdot y^{3}-y\cdot \sin x)\cdot \frac{dx}{dt} + (3\cdot x^{2}\cdot y^{2}+\cos x)\cdot \frac{dy}{dt}[/tex]
[tex]x = \ln t^{2}[/tex]
[tex]x = 2\cdot \ln t[/tex]
[tex]\frac{dx}{dt} = \frac{2}{t}[/tex], [tex]\frac{dy}{dt} = 4\cdot \cos 4t[/tex]
[tex]\frac{dz}{dt} = [4\cdot \ln t\cdot \sin^{3} 4t - \sin 4t \cdot \sin (2\ln t)]\cdot \left(\frac{2}{t} \right)+[12\cdot (\ln t)^{2}\cdot \sin^{2} 4t + \cos (2\cdot \ln t)]\cdot (4\cdot \cos 4t)[/tex]
