Let $f(x, y, z)=y^{2} z$ and $g(t)=\left(t, t^{2}, t^{3}\right)$.
$$
\begin{array}{l}
h(t)=f(g(t)) \
h^{\prime}(2)=
\end{array}
$$

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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $h'(t)$, we need to apply the chain rule to the composite function $h(t) = f(g(t))$. ‖ ‖ step 2 ⋮ The chain rule states that if $h(t) = f(g(t))$, then $h'(t) = f'(g(t)) \cdot g'(t)$. ‖ ‖ step 3 ⋮ We have $f(x, y, z) = y^2z$ and $g(t) = (t, t^2, t^3)$. Thus, $g'(t) = (1, 2t, 3t^2)$. ‖ ‖ step 4 ⋮ To find $f'(g(t))$, we differentiate $f(x, y, z)$ with respect to $x$, $y$, and $z$ and evaluate at $g(t)$. ‖ ‖ step 5 ⋮ We get $\frac{\partial f}{\partial x} = 0$, $\frac{\partial f}{\partial y} = 2yz$, and $\frac{\partial f}{\partial z} = y^2$. ‖ ‖ step 6 ⋮ Evaluating these derivatives at $g(t)$ gives $\frac{\partial f}{\partial y}(g(t)) = 2t^2 \cdot t^3 = 2t^5$ and $\frac{\partial f}{\partial z}(g(t)) = (t^2)^2 = t^4$. ‖ ‖ step 7 ⋮ Now, $f'(g(t)) = \left(0, 2t^5, t^4\right)$. ‖ ‖ step 8 ⋮ Multiplying $f'(g(t))$ by $g'(t)$ component-wise gives $h'(t) = \left(0, 2t^5, t^4\right) \cdot (1, 2t, 3t^2)$. ‖ ‖ step 9 ⋮ This results in $h'(t) = 0 + 4t^6 + 3t^6 = 7t^6$. ‖ ‖ step 10 ⋮ Finally, evaluate $h'(t)$ at $t = 2$ to find $h'(2) = 7(2)^6$. ‖ ‖ step 11 ⋮ Calculating the value gives $h'(2) = 7 \cdot 64 = 448$. ‖ ∻[question number] Answer∻ ⚹ $h'(2) = 448$ ⚹ ∻Key Concept∻ ⚹Chain Rule for Differentiation of Composite Functions⚹ ∻Explanation∻ ⚹To find the derivative of a composite function, we apply the chain rule, which involves taking the derivative of the outer function evaluated at the inner function and multiplying it by the derivative of the inner function.⚹◊What is the formula for finding the derivative of a composite function at a specific value?⍭ Generate me a similar question◊

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