Question

Math

Posted 2 months ago

```
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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Answer from Sia

Posted 2 months ago

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.

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