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The problem requires finding the derivative \( f'(x) \) for each of the following four cases. Let me go through each one step by step:

### 1) \( f(x) = \sin(x^2 + x + 1) \)
We need to differentiate using the chain rule. The outer function is \( \sin(u) \), where \( u = x^2 + x + 1 \).

\[
f'(x) = \cos(x^2 + x + 1) \cdot \frac{d}{dx}(x^2 + x + 1)
\]

\[
f'(x) = \cos(x^2 + x + 1) \cdot (2x + 1)
\]

Thus, the derivative is:
\[
f'(x) = (2x + 1) \cdot \cos(x^2 + x + 1)
\]

---

### 2) \( f(x) = \sqrt[3]{x^2 + 3x - 4} \)
We will apply the chain rule, with the outer function being the cube root, which can be written as \( (x^2 + 3x - 4)^{1/3} \).

\[
f'(x) = \frac{1}{3}(x^2 + 3x - 4)^{-2/3} \cdot \frac{d}{dx}(x^2 + 3x - 4)
\]

\[
f'(x) = \frac{1}{3}(x^2 + 3x - 4)^{-2/3} \cdot (2x + 3)
\]

Thus, the derivative is:
\[
f'(x) = \frac{2x + 3}{3(x^2 + 3x - 4)^{2/3}}
\]

---

### 3) \( f(x) = \frac{\sqrt{x^2 + x}}{\sqrt{x - 1}} \)
This requires using both the quotient rule and the chain rule. Start by simplifying the problem:

\[
f(x) = \frac{\sqrt{x^2 + x}}{\sqrt{x - 1}} = \frac{(x^2 + x)^{1/2}}{(x - 1)^{1/2}}
\]

We can use the quotient rule, where \( u = (x^2 + x)^{1/2} \) and \( v = (x - 1)^{1/2} \).

The quotient rule is:
\[
f'(x) = \frac{v u' - u v'}{v^2}
\]

Where:
- \( u' = \frac{1}{2}(x^2 + x)^{-1/2} \cdot (2x + 1) = \frac{(2x + 1)}{2\sqrt{x^2 + x}} \)
- \( v' = \frac{1}{2}(x - 1)^{-1/2} \)

Thus:
\[
f'(x) = \frac{\sqrt{x - 1} \cdot \frac{2x + 1}{2\sqrt{x^2 + x}} - \sqrt{x^2 + x} \cdot \frac{1}{2\sqrt{x - 1}}}{x - 1}
\]

Simplifying this expression gives the final result for the derivative.

---

### 4) \( f(x) = \frac{4\sqrt[3]{x}}{x\sqrt{x + 1}} \)
Here, we again use the quotient rule. Let \( u = 4x^{1/3} \) and \( v = x\sqrt{x + 1} = x(x + 1)^{1/2} \).

The quotient rule is:
\[
f'(x) = \frac{v u' - u v'}{v^2}
\]

Where:
- \( u' = \frac{4}{3}x^{-2/3} \)
- \( v' = \sqrt{x + 1} + \frac{x}{2\sqrt{x + 1}} \)

Thus:
\[
f'(x) = \frac{x(x + 1)^{1/2} \cdot \frac{4}{3}x^{-2/3} - 4x^{1/3} \cdot \left( \sqrt{x + 1} + \frac{x}{2\sqrt{x + 1}} \right)}{x^2(x + 1)}
\]

Simplifying will give the final expression for the derivative.

---

Would you like further details on the simplification for any of these derivatives?

Here are 5 related questions to explore:
1. How does the quotient rule apply to more complex fractions?
2. What are the common mistakes to avoid when applying the chain rule?
3. Can the product and chain rules be combined to solve more advanced problems?
4. How would the derivatives change if higher-order derivatives were needed?
5. What are the general techniques for simplifying complex rational functions?

**Tip**: When using the chain rule, always differentiate the outer function first, then multiply by the derivative of the inner function.

Calculate the derivative f'(x) for the following functions:
1) f(x) = sin(x^2 + x + 1)
2) f(x) = (x^2 + 3x - 4)^(1/3)
3) f(x) = sqrt(x^2 + x) / sqrt(x - 1)
4) f(x) = 4*sqrt[3]{x} / (x*sqrt(x + 1))

Learn how to differentiate complex functions using the chain and quotient rules. This problem involves calculating derivatives for trigonometric, radical, and rational functions, suitable for advanced calculus students.

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