Question

SENDO Z(X,Y)=e~x+3y ,x(t)=t² e y (t)= t³+1 entao dz/dt e:

Answer 1

pela regra da cadeia
$\boxed{\boxed{\frac{dZ}{dt} =\left( \frac{\partial Z}{\partial x} * \frac{\partial x}{\partial t} \right) + \left(\frac{\partial Z}{\partial y} * \frac{\partial y}{\partial t} \right)}}$

temos:

$\bmatrix Z=e^x +3y\\\\ \frac{\partial Z}{\partial x} = e^x +0 \\\\\frac{\partial Z}{\partial y}=0+3*1 \\\\\\ x=t^2\\\\ \frac{\partial x}{\partial t}= 2t \\\\\\ y=t^3+1\\\\ \frac{\partial y}{\partial t}=3t \end$

substituindo
$\frac{\partial Z}{\partial x} = e^x*2t +3*3t^2\\\\ \frac{\partial Z}{\partial x} = e^x*2t + 9t^2\\\\ \boxed{\boxed{\frac{\partial Z}{\partial x} = e^{t^2}*2t+ 9t^2}}$