To find the derivative of $h(x) = e^{x^{4 - x + 1}}$, let us proceed step by step:

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Step 1: Write down the function.

The function is given as: $h(x) = e^{x^{4 - x + 1}}$

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Step 2: Apply the chain rule.

The chain rule states: $\frac{d}{dx} \big( e^u \big) = e^u \cdot \frac{du}{dx}$

Here, $u = x^{4 - x + 1}$. So, the derivative is: $h'(x) = e^{x^{4 - x + 1}} \cdot \frac{d}{dx} \big( x^{4 - x + 1} \big)$

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Step 3: Differentiate $u = x^{4 - x + 1}$.

To differentiate $u = x^{4 - x + 1}$, we use the general power rule for derivatives of $x^v$, which states: $\frac{d}{dx} \big( x^v \big) = x^v \cdot \big( \ln(x) \cdot \frac{dv}{dx} + \frac{v}{x} \big)$

Here, $v = 4 - x + 1 = 5 - x$. Therefore: $\frac{d}{dx} \big( x^{4 - x + 1} \big) = x^{5 - x} \cdot \Big( \ln(x) \cdot \frac{d}{dx}(-x + 5) + \frac{5 - x}{x} \Big)$

1. $\frac{d}{dx}(-x + 5) = -1$
2. Substitute everything into the expression: $\frac{d}{dx} \big( x^{4 - x + 1} \big) = x^{5 - x} \cdot \Big( \ln(x) \cdot (-1) + \frac{5 - x}{x} \Big)$

Simplify the terms: $\frac{d}{dx} \big( x^{4 - x + 1} \big) = x^{5 - x} \cdot \Big( \frac{5 - x}{x} - \ln(x) \Big)$

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Step 4: Substitute back into the derivative of $h(x)$.

Now, substitute back into $h'(x)$: $h'(x) = e^{x^{5 - x}} \cdot x^{5 - x} \cdot \Big( \frac{5 - x}{x} - \ln(x) \Big)$

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Final Answer:

$h'(x) = e^{x^{5 - x}} \cdot x^{5 - x} \cdot \Big( \frac{5 - x}{x} - \ln(x) \Big)$

Let me know if you would like further clarifications or additional examples.

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Relative Questions:

1. How does the chain rule apply to exponential functions like $e^u$?
2. What are the conditions for differentiating powers like $x^v$, where $v$ is a variable?
3. Can you show how to simplify further when $x > 0$?
4. What is the interpretation of this derivative in terms of rate of change?
5. How do logarithmic functions play a role in differentiation?

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Tip:

When differentiating $x^v$ where $v$ depends on $x$, always remember to combine the product rule with the natural logarithm chain rule for precise results.