Suppose we have a vector-valued function $g(t)$ and a scalar function $f(x, y)$. Let $h(t)=f(g(t))$.

We know:
$$
\begin{array}{l}
g(2)=(-4,2) \
g^{\prime}(2)=(1,3) \
\nabla f(-4,2)=(6,-2)
\end{array}
$$

Evaluate $\frac{d h}{d t}$ at $t=2$.
$$
h^{\prime}(2)=
$$

Question

Suppose we have a vector-valued function $g(t)$ and a scalar function $f(x, y)$. Let $h(t)=f(g(t))$.

We know:
$$
\begin{array}{l}
g(2)=(-4,2) \
g^{\prime}(2)=(1,3) \
\nabla f(-4,2)=(6,-2)
\end{array}
$$

Evaluate $\frac{d h}{d t}$ at $t=2$.
$$
h^{\prime}(2)=
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the derivative of the composite function $h(t) = f(g(t))$, we use the chain rule. ‖ ‖ step 2 ⋮ The chain rule states that $h'(t) = abla f(g(t)) \cdot g'(t)$. ‖ ‖ step 3 ⋮ We substitute the given values into the chain rule formula: $ abla f(-4,2) = (6,-2)$ and $g'(2) = (1,3)$. ‖ ‖ step 4 ⋮ We calculate the dot product of $ abla f(-4,2)$ and $g'(2)$: $(6,-2) \cdot (1,3) = 6*1 + (-2)*3$. ‖ ‖ step 5 ⋮ Simplifying the dot product: $6*1 + (-2)*3 = 6 - 6 = 0$. ‖ ∻Answer∻ ⚹ $h'(2) = 0$ ⚹ ∻Key Concept∻ ⚹Chain Rule for Derivatives⚹ ∻Explanation∻ ⚹The derivative of a composite function $h(t) = f(g(t))$ at a point is found by taking the dot product of the gradient of $f$ at $g(t)$ and the derivative of $g$ at $t$.⚹◊What is the chain rule in calculus and how does it apply to finding the derivative of h(t) = f(g(t))?⍭ Generate me a similar question◊

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