	Find the derivative f^' for each function.  Simply your answer.

	f(x)=x^(2⁄3)-x^(1⁄3)+4

	f(x)=(3x-2x^2 )(5+4x)

	f(x)=e^x  sin⁡x

	f(x)=(3-2x-x^2)/(x^2-1)

	f(x)=x cos⁡x+sin⁡x

	f(x)=(2x+5)/√x


Let's find the derivative \( f'(x) \) for each of the given functions and simplify the answers.

### 1. \( f(x) = x^{\frac{2}{3}} - x^{\frac{1}{3}} + 4 \)

To differentiate this, apply the power rule:

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

### 2. \( f(x) = (3x - 2x^2)(5 + 4x) \)

Use the product rule: \( (uv)' = u'v + uv' \).

Let \( u = 3x - 2x^2 \) and \( v = 5 + 4x \).

Differentiate \( u \) and \( v \):

\[
u' = 3 - 4x
\]
\[
v' = 4
\]

Now apply the product rule:

\[
f'(x) = (3 - 4x)(5 + 4x) + (3x - 2x^2)(4)
\]
\[
f'(x) = (15 + 12x - 20x - 16x^2) + (12x - 8x^2)
\]
\[
f'(x) = -24x^2 - 8x + 15
\]

### 3. \( f(x) = e^x \sin x \)

Use the product rule:

Let \( u = e^x \) and \( v = \sin x \).

Differentiate \( u \) and \( v \):

\[
u' = e^x
\]
\[
v' = \cos x
\]

Apply the product rule:

\[
f'(x) = e^x \cos x + e^x \sin x
\]
\[
f'(x) = e^x (\cos x + \sin x)
\]

### 4. \( f(x) = \frac{3 - 2x - x^2}{x^2 - 1} \)

Use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

Let \( u = 3 - 2x - x^2 \) and \( v = x^2 - 1 \).

Differentiate \( u \) and \( v \):

\[
u' = -2 - 2x
\]
\[
v' = 2x
\]

Now apply the quotient rule:

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

Simplify the expression:

\[
f'(x) = \frac{-2x^2 + 2 - 2x^3 + 2x - (6x - 4x^2 - 2x^3)}{(x^2 - 1)^2}
\]
\[
f'(x) = \frac{-4x^3 + 2x^2 - 4x + 2}{(x^2 - 1)^2}
\]

### 5. \( f(x) = x \cos x + \sin x \)

Use the product rule for the first term:

Let \( u = x \) and \( v = \cos x \).

Differentiate \( u \) and \( v \):

\[
u' = 1
\]
\[
v' = -\sin x
\]

Now apply the product rule:

\[
f'(x) = 1 \cdot \cos x + x(-\sin x) + \cos x
\]
\[
f'(x) = \cos x - x \sin x
\]

### 6. \( f(x) = \frac{2x + 5}{\sqrt{x}} \)

This is a quotient. Simplify \( f(x) \) first:

\[
f(x) = (2x + 5)x^{-\frac{1}{2}}
\]

Now use the product rule:

Let \( u = 2x + 5 \) and \( v = x^{-\frac{1}{2}} \).

Differentiate \( u \) and \( v \):

\[
u' = 2
\]
\[
v' = -\frac{1}{2}x^{-\frac{3}{2}}
\]

Apply the product rule:

\[
f'(x) = 2x^{-\frac{1}{2}} + (2x + 5)\left(-\frac{1}{2}x^{-\frac{3}{2}}\right)
\]
\[
f'(x) = \frac{2}{\sqrt{x}} - \frac{2x + 5}{2x^{\frac{3}{2}}}
\]

Would you like further details on any of these solutions, or do you have any questions?

---

**5 Related Questions:**

1. How does the power rule apply to fractional exponents?
2. What is the importance of the product rule in differentiation?
3. How do you simplify expressions after applying the quotient rule?
4. How do you handle derivatives of trigonometric functions?
5. What are the common mistakes to avoid in differentiation?

**Tip:** Always simplify the function before differentiating if possible—it often makes the process easier and the derivative simpler.

Find the derivative f^' for each function. Simplify your answer.

f(x)=x^(2⁄3)-x^(1⁄3)+4

f(x)=(3x-2x^2)(5+4x)

f(x)=e^x  sin⁡x

f(x)=(3-2x-x^2)/(x^2-1)

f(x)=x cos⁡x+sin⁡x

f(x)=(2x+5)/√x

Solve complex differentiation problems involving power, product, and quotient rules. Learn how to differentiate functions such as f(x) = x^(2/3) - x^(1/3) + 4 and f(x) = (3x - 2x^2)(5 + 4x). Suitable for advanced calculus students in Grades 11-12.

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