Consider the following. $$ w=x y^{2}+x^{2} z+y z^{2}, \quad x=t^{2}, \quad y=2 t, \quad z=2 $$ (a) Find $$ d w / d t $$ using the appropriate Chain Rule. $$ \frac{d w}{d t}=\square $$ (b) Find $$ d w / d t $$ by converting $$ w $$ to a function of $$ t $$ before differentiating. $$ \frac{d w}{d t}=\square $$

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We are given
$$
w = xy^2 + x^2z + yz^2,\quad x = t^2,\quad y = 2t,\quad z = 2.
$$

### (a) Using the Chain Rule

We start with the chain rule:
$$
\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}.
$$

**Step 1. Compute the partial derivatives of $$ w $$.**

1. The partial derivative with respect to $$ x $$:
   $$
   \frac{\partial w}{\partial x} = \frac{\partial}{\partial x}\left(xy^2\right) + \frac{\partial}{\partial x}\left(x^2z\right) + \frac{\partial}{\partial x}\left(yz^2\right) = y^2 + 2xz + 0 = y^2 + 2xz.
   $$

2. The partial derivative with respect to $$ y $$:
   $$
   \frac{\partial w}{\partial y} = \frac{\partial}{\partial y}\left(xy^2\right) + \frac{\partial}{\partial y}\left(x^2z\right) + \frac{\partial}{\partial y}\left(yz^2\right) = 2xy + 0 + z^2 = 2xy + z^2.
   $$

3. The partial derivative with respect to $$ z $$:
   $$
   \frac{\partial w}{\partial z} = \frac{\partial}{\partial z}\left(xy^2\right) + \frac{\partial}{\partial z}\left(x^2z\right) + \frac{\partial}{\partial z}\left(yz^2\right) = 0 + x^2 + 2yz = x^2 + 2yz.
   $$

**Step 2. Compute the derivatives of $$ x $$, $$ y $$, and $$ z $$ with respect to $$ t $$.**

- $$ x = t^2 $$ gives
  $$
  \frac{dx}{dt} = 2t.
  $$
- $$ y = 2t $$ gives
  $$
  \frac{dy}{dt} = 2.
  $$
- $$ z = 2 $$ gives
  $$
  \frac{dz}{dt} = 0.
  $$

**Step 3. Substitute into the chain rule formula.**

$$
\frac{dw}{dt} = \left(y^2+2xz\right)(2t) + \left(2xy+z^2\right)(2) + \left(x^2+2yz\right)(0).
$$

Since the last term vanishes, we have:
$$
\frac{dw}{dt} = \left(y^2+2xz\right)(2t) + \left(2xy+z^2\right)(2).
$$

**Step 4. Substitute the given values $$ x=t^2,\, y=2t,\, z=2 $$.**

- Calculate $$ y^2 $$:
  $$
  (2t)^2 = 4t^2.
  $$
- Calculate $$ 2xz $$:
  $$
  2(t^2)(2) = 4t^2.
  $$
- So, $$ y^2+2xz = 4t^2+4t^2 = 8t^2 $$.

- Next, calculate $$ 2xy $$:
  $$
  2(t^2)(2t) = 4t^3.
  $$
- And $$ z^2 $$:
  $$
  2^2 = 4.
  $$
- So, $$ 2xy+z^2 = 4t^3 + 4 $$.

Now, substitute back:
$$
\frac{dw}{dt} = \left(8t^2\right)(2t) + \left(4t^3+4\right)(2) = 16t^3 + 8t^3 + 8.
$$
$$
\frac{dw}{dt} = 24t^3 + 8.
$$

Thus, for part (a),
$$
\frac{dw}{dt} = 24t^3+8.
$$

### (b) Converting $$ w $$ to a Function of $$ t $$ First

**Step 1. Substitute $$ x = t^2 $$, $$ y = 2t $$, and $$ z = 2 $$ into $$ w $$.**

$$
w = (t^2)(2t)^2 + (t^2)^2(2) + (2t)(2)^2.
$$

**Step 2. Simplify each term.**

1. First term:
   $$
   (t^2)(2t)^2 = t^2(4t^2) = 4t^4.
   $$
2. Second term:
   $$
   (t^2)^2(2) = t^4\cdot 2 = 2t^4.
   $$
3. Third term:
   $$
   (2t)(2)^2 = 2t\cdot 4 = 8t.
   $$

Combine them:
$$
w = 4t^4 + 2t^4 + 8t = 6t^4 + 8t.
$$

**Step 3. Differentiate $$ w(t) $$ with respect to $$ t $$.**

$$
\frac{dw}{dt} = \frac{d}{dt}(6t^4 + 8t) = 24t^3 + 8.
$$

Thus, for part (b),
$$
\frac{dw}{dt} = 24t^3+8.
$$
For part (a), $$ \frac{dw}{dt} = 24t^3 + 8 $$. For part (b), $$ \frac{dw}{dt} = 24t^3 + 8 $$.

Consider the following.

(a) Find using the appropriate Chain Rule.

(b) Find by converting to a function of before differentiating.

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The Deep Dive

(a) Using the Chain Rule:

(b) Converting to a function of :

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Consider the following. (a) Find using the appropriate Chain Rule. (b) Find by converting to a function of before differentiating.