We need to find $\frac{dw}{dt}$ for each of the given functions using the chain rule, and express the final answer in terms of $t$.

1. Problem Description

We need to find $\frac{dw}{dt}$ for each of the given functions using the chain rule, and express the final answer in terms of $t$.

2. Solution Steps

Problem 1: $w = x^2y^3$, $x = t^3$, $y = t^2$

We use the chain rule:

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt}$

First, we find the partial derivatives:

$\frac{\partial w}{\partial x} = 2xy^3$

$\frac{\partial w}{\partial y} = 3x^2y^2$

Next, we find the derivatives of $x$ and $y$ with respect to $t$:

$\frac{dx}{dt} = 3t^2$

$\frac{dy}{dt} = 2t$

Now, we substitute these into the chain rule formula:

$\frac{dw}{dt} = (2xy^3)(3t^2) + (3x^2y^2)(2t)$

Substitute $x = t^3$ and $y = t^2$:

$\frac{dw}{dt} = (2(t^3)(t^2)^3)(3t^2) + (3(t^3)^2(t^2)^2)(2t)$

$\frac{dw}{dt} = (2t^3t^6)(3t^2) + (3t^6t^4)(2t)$

$\frac{dw}{dt} = (2t^9)(3t^2) + (3t^{10})(2t)$

$\frac{dw}{dt} = 6t^{11} + 6t^{11}$

$\frac{dw}{dt} = 12t^{11}$

Problem 2: $w = x^2y - y^2x$, $x = \cos t$, $y = \sin t$

We use the chain rule:

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt}$

First, we find the partial derivatives:

$\frac{\partial w}{\partial x} = 2xy - y^2$

$\frac{\partial w}{\partial y} = x^2 - 2yx$

Next, we find the derivatives of $x$ and $y$ with respect to $t$:

$\frac{dx}{dt} = -\sin t$

$\frac{dy}{dt} = \cos t$

Now, we substitute these into the chain rule formula:

$\frac{dw}{dt} = (2xy - y^2)(-\sin t) + (x^2 - 2yx)(\cos t)$

Substitute $x = \cos t$ and $y = \sin t$:

$\frac{dw}{dt} = (2\cos t \sin t - \sin^2 t)(-\sin t) + (\cos^2 t - 2\sin t \cos t)(\cos t)$

$\frac{dw}{dt} = -2\cos t \sin^2 t + \sin^3 t + \cos^3 t - 2\sin t \cos^2 t$

$\frac{dw}{dt} = \cos^3 t - 2\cos t \sin^2 t - 2\sin t \cos^2 t + \sin^3 t$

$\frac{dw}{dt} = \cos^3 t + \sin^3 t - 2\cos t \sin t(\sin t + \cos t)$

$\frac{dw}{dt} = (\cos t + \sin t)(\cos^2 t - \cos t \sin t + \sin^2 t) - 2\cos t \sin t (\cos t + \sin t)$

$\frac{dw}{dt} = (\cos t + \sin t)(1 - \cos t \sin t) - 2\cos t \sin t(\cos t + \sin t)$

$\frac{dw}{dt} = (\cos t + \sin t)(1 - \cos t \sin t - 2\cos t \sin t)$

$\frac{dw}{dt} = (\cos t + \sin t)(1 - 3\cos t \sin t)$

Problem 3: $w = e^x\sin y + e^y\sin x$, $x = 3t$, $y = 2t$

We use the chain rule:

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt}$

First, we find the partial derivatives:

$\frac{\partial w}{\partial x} = e^x\sin y + e^y\cos x$

$\frac{\partial w}{\partial y} = e^x\cos y + e^y\sin x$

Next, we find the derivatives of $x$ and $y$ with respect to $t$:

$\frac{dx}{dt} = 3$

$\frac{dy}{dt} = 2$

Now, we substitute these into the chain rule formula:

$\frac{dw}{dt} = (e^x\sin y + e^y\cos x)(3) + (e^x\cos y + e^y\sin x)(2)$

Substitute $x = 3t$ and $y = 2t$:

$\frac{dw}{dt} = 3(e^{3t}\sin(2t) + e^{2t}\cos(3t)) + 2(e^{3t}\cos(2t) + e^{2t}\sin(3t))$

$\frac{dw}{dt} = 3e^{3t}\sin(2t) + 3e^{2t}\cos(3t) + 2e^{3t}\cos(2t) + 2e^{2t}\sin(3t)$

$\frac{dw}{dt} = e^{3t}(3\sin(2t) + 2\cos(2t)) + e^{2t}(3\cos(3t) + 2\sin(3t))$

Problem 4: $w = \ln(x/y)$, $x = \tan t$, $y = \sec^2 t$

$w = \ln(x) - \ln(y)$

We use the chain rule:

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt}$

First, we find the partial derivatives:

$\frac{\partial w}{\partial x} = \frac{1}{x}$

$\frac{\partial w}{\partial y} = -\frac{1}{y}$

Next, we find the derivatives of $x$ and $y$ with respect to $t$:

$\frac{dx}{dt} = \sec^2 t$

$\frac{dy}{dt} = 2\sec t (\sec t \tan t) = 2\sec^2 t \tan t$

Now, we substitute these into the chain rule formula:

$\frac{dw}{dt} = (\frac{1}{x})(\sec^2 t) + (-\frac{1}{y})(2\sec^2 t \tan t)$

Substitute $x = \tan t$ and $y = \sec^2 t$:

$\frac{dw}{dt} = (\frac{1}{\tan t})(\sec^2 t) + (-\frac{1}{\sec^2 t})(2\sec^2 t \tan t)$

$\frac{dw}{dt} = \frac{\sec^2 t}{\tan t} - 2\tan t$

$\frac{dw}{dt} = \frac{1}{\cos^2 t} \cdot \frac{\cos t}{\sin t} - 2\tan t$

$\frac{dw}{dt} = \frac{1}{\cos t \sin t} - 2\tan t$

$\frac{dw}{dt} = \frac{1}{\cos t \sin t} - \frac{2\sin t}{\cos t}$

$\frac{dw}{dt} = \frac{1 - 2\sin^2 t}{\cos t \sin t}$

$\frac{dw}{dt} = \frac{\cos 2t}{\frac{1}{2}\sin 2t} = 2\cot 2t$

Problem 5: $w = \sin(xyz^2)$, $x = t^3$, $y = t^2$, $z = t$

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$

$\frac{\partial w}{\partial x} = \cos(xyz^2)yz^2$

$\frac{\partial w}{\partial y} = \cos(xyz^2)xz^2$

$\frac{\partial w}{\partial z} = \cos(xyz^2)2xyz$

$\frac{dx}{dt} = 3t^2$

$\frac{dy}{dt} = 2t$

$\frac{dz}{dt} = 1$

$\frac{dw}{dt} = \cos(xyz^2)[yz^2(3t^2) + xz^2(2t) + 2xyz(1)]$

$\frac{dw}{dt} = \cos((t^3)(t^2)(t)^2)[(t^2)(t)^2(3t^2) + (t^3)(t)^2(2t) + 2(t^3)(t^2)(t)(1)]$

$\frac{dw}{dt} = \cos(t^7)[3t^6 + 2t^6 + 2t^6]$

$\frac{dw}{dt} = \cos(t^7)[7t^6]$

$\frac{dw}{dt} = 7t^6\cos(t^7)$

Problem 6: $w = xy + yz + xz$, $x = t^2$, $y = 1-t^2$, $z = 1-t$

$\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$

$\frac{\partial w}{\partial x} = y + z$

$\frac{\partial w}{\partial y} = x + z$

$\frac{\partial w}{\partial z} = y + x$

$\frac{dx}{dt} = 2t$

$\frac{dy}{dt} = -2t$

$\frac{dz}{dt} = -1$

$\frac{dw}{dt} = (y+z)(2t) + (x+z)(-2t) + (x+y)(-1)$

$\frac{dw}{dt} = (1-t^2 + 1-t)(2t) + (t^2 + 1-t)(-2t) + (t^2 + 1-t^2)(-1)$

$\frac{dw}{dt} = (2-t-t^2)(2t) + (t^2-t+1)(-2t) + (1)(-1)$

$\frac{dw}{dt} = 4t - 2t^2 - 2t^3 - 2t^3 + 2t^2 - 2t - 1$

$\frac{dw}{dt} = 2t - 4t^3 - 1$

$\frac{dw}{dt} = -4t^3 + 2t - 1$

3. Final Answer

1. $\frac{dw}{dt} = 12t^{11}$

2. $\frac{dw}{dt} = (\cos t + \sin t)(1 - 3\cos t \sin t)$

3. $\frac{dw}{dt} = e^{3t}(3\sin(2t) + 2\cos(2t)) + e^{2t}(3\cos(3t) + 2\sin(3t))$

4. $\frac{dw}{dt} = 2\cot 2t$

5. $\frac{dw}{dt} = 7t^6\cos(t^7)$

6. $\frac{dw}{dt} = -4t^3 + 2t - 1$