---
# System prepended metadata

title: In-Class Activity 3.3
tags: ['3.3']

---

[toc]

# Question 1

### (a) Simplify the following expression:

$$
(2x^2+1)'(3x^2-1) - (2x^2+1)(3x^2-1)' = \quad \text{______}
$$

::: spoiler
<summary> Example: </summary>

# Example: Simplify the following expression

$$
(3x^4+1)'(5x^3-1) - (3x^4+1)(5x^3-1)'
$$

---
\begin{align*}
&(3x^4+1)'(5x^3-1) - (3x^4+1)(5x^3-1)' \\
&= (12x^3)(5x^3 - 1) - (3x^4 + 1)(15x^2) \\
&= 12x^3(5x^3) - 12x^3(1) - (3x^4)(15x^2) - (1)(15x^2) \\
&= 60x^6 - 12x^3 - 45x^6 - 15x^2 \\
&= (60x^6 - 45x^6) - 12x^3 - 15x^2 \\
&= 15x^6 - 12x^3 - 15x^2
\end{align*}


---

### Final Answer:

$$
(3x^4+1)'(5x^3-1) - (3x^4+1)(5x^3-1)' = 15x^6 - 12x^3 - 15x^2
$$



:::

---

### (b) Simplify the following expression:

$$
(x^4)'(4x+6) - x^4(4x+6)' = \quad \text{______}
$$

::: spoiler
<summary> Example: </summary>

We are asked to differentiate the expression:

$$
(x^3)'(5x+7) - x^3(5x+7)'
$$


Now let's break it down step by step.

### Step 1: Differentiate $(x^3)'$

The derivative of $x^3$ is:

$$
(x^3)' = 3x^2
$$

### Step 2: Differentiate $(5x + 7)'$

The derivative of $5x + 7$ is:

$$
(5x + 7)' = 5
$$

### Step 3: Substitute into the expression

Now substitute the derivatives back into the original expression:

$$
(x^3)'(5x + 7) - x^3(5x + 7)' = 3x^2(5x + 7) - x^3(5)
$$

### Step 4: Simplify

Now simplify the expression:

$$
3x^2(5x + 7) - 5x^3 = 3x^2(5x) + 3x^2(7) - 5x^3
$$

$$
= 15x^3 + 21x^2 - 5x^3
$$

Now combine like terms:

$$
(15x^3 - 5x^3) + 21x^2 = 10x^3 + 21x^2
$$

### Final Answer:

The simplified expression is:

$$
10x^3 + 21x^2
$$


:::

---

### (c) Simplify the following expression:

$$
(2x+1)'(6x-4) - (2x+1)(6x-4)' = \quad \text{______}
$$

::: spoiler
<summary> Example: </summary>

We are asked to differentiate the expression:

$$
(3x+2)'(5x-7) - (3x+2)(5x-7)'
$$

This is another application of the product rule. Let's go through the steps.

### Step 1: Differentiate $(3x+2)'$

The derivative of $3x + 2$ is:

$$
(3x+2)' = 3
$$

### Step 2: Differentiate $(5x-7)'$

The derivative of $5x - 7$ is:

$$
(5x-7)' = 5
$$

### Step 3: Substitute into the expression

Now substitute the derivatives back into the original expression:

$$
(3x+2)'(5x-7) - (3x+2)(5x-7)' = 3(5x-7) - (3x+2)(5)
$$

### Step 4: Simplify

Now simplify the expression:

$$
3(5x - 7) - (3x + 2)(5) = 3(5x) - 3(7) - 5(3x) - 5(2)
$$

$$
= 15x - 21 - 15x - 10
$$

Now combine like terms:

$$
15x - 15x - 21 - 10 = -31
$$

### Final Answer:

The simplified expression is:

$$
-31
$$


:::

---

### (d) Simplify the following expression:

$$
(5x-4)'(3x+7) - (5x-4)(3x+7)' = \quad \text{______}
$$

::: spoiler
<summary> Example: </summary>

We are asked to differentiate the expression:

$$
(5x-4)'(3x+7) - (5x-4)(3x+7)'
$$

This is another application of the product rule. Let's break it down.

### Step 1: Differentiate $(5x-4)'$

The derivative of $5x - 4$ is:

$$
(5x-4)' = 5
$$

### Step 2: Differentiate $(3x+7)'$

The derivative of $3x + 7$ is:

$$
(3x+7)' = 3
$$

### Step 3: Substitute into the expression

Now substitute the derivatives back into the original expression:

$$
(5x-4)'(3x+7) - (5x-4)(3x+7)' = 5(3x+7) - (5x-4)(3)
$$

### Step 4: Simplify

Now simplify the expression:

$$
5(3x + 7) - 3(5x - 4) = 5(3x) + 5(7) - 3(5x) + 3(4)
$$

$$
= 15x + 35 - 15x + 12
$$

Now combine like terms:

$$
15x - 15x + 35 + 12 = 47
$$

### Final Answer:

The simplified expression is:

$$
47
$$


:::

---

# Question 2

### (a) Find $f'(x)$ for the following and simplify:

$$
f(x) = x^2 e^x \quad \Rightarrow \quad f'(x) = \text{______}
$$

::: spoiler
<summary> Example: </summary>

Use the product rule to differentiate $f(x) = x^3 e^x$.

| Function     | $x^3$    | $e^x$  |
|--------------|--------------|------------|
| Derivative   | $3x^2$   | $e^x$  |

---

### Apply the Product Rule:

The product rule states:

$$
(fg)' = f'g + fg'
$$

For $f(x) = x^3 e^x$:

$$
f'(x) = (3x^2)(e^x) + (x^3)(e^x)
$$

---

### Final Answer:

$$
f'(x) = 3x^2 e^x + x^3 e^x
$$

:::

---

### (b) Find $f'(x)$ for the following and simplify:

$$
f(x) = x^3 \ln(x) \quad \Rightarrow \quad f'(x) = \text{______}
$$

::: spoiler
<summary> Example: </summary>

Use the product rule to differentiate $f(x) = x^4 \ln(x)$.

| Function     | $x^4$    | $\ln(x)$    |
|--------------|----------|-------------|
| Derivative   | $4x^3$   | $\frac{1}{x}$  |

---

### Apply the Product Rule:

The product rule states:

$$
(fg)' = f'g + fg'
$$

For $f(x) = x^4 \ln(x)$:

$$
f'(x) = (4x^3)(\ln(x)) + (x^4)\left(\frac{1}{x}\right)
$$

---

### Final Answer:

$$
f'(x) = 4x^3 \ln(x) + x^3
$$


:::

---

# Question 3

### (a) Find $f'(x)$ and simplify for:

$$
f(x) = \frac{2x+3}{x-2} \quad \Rightarrow \quad f'(x) = \text{______}
$$

::: spoiler
<summary> Example: </summary>

Use the quotient rule to differentiate $f(x) = \dfrac{5x+4}{x-3}$.

| Function     | $5x+4$    | $x-3$    |
|--------------|-----------|----------|
| Derivative   | $5$       | $1$      |

---

### Apply the Quotient Rule:

The quotient rule states:

$$
\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}
$$

For $f(x) = \frac{5x+4}{x-3}$:

$$
f'(x) = \frac{(5)(x-3) - (5x+4)(1)}{(x-3)^2}
$$

---

### Detailed Simplification:

1. Distribute in the first term: $(5)(x-3)$
   $$
   5(x-3) = 5x - 15
   $$

2. Expand the second term: $(5x+4)(1)$
   $$
   5x+4
   $$

3. Combine the terms in the numerator:
   $$
   5x - 15 - (5x + 4)
   $$

4. Distribute the negative sign:
   $$
   5x - 15 - 5x - 4 = -19
   $$

So the expression becomes:

$$
f'(x) = \frac{-19}{(x-3)^2}
$$

---

### Final Answer:

$$
f'(x) = \frac{-19}{(x-3)^2}
$$


:::

---

### (b) Find $f'(x)$ and simplify for:

$$
f(x) = \frac{x^2+2}{x^2-3} \quad \Rightarrow \quad f'(x) = \text{______}
$$

::: spoiler
<summary> Example: </summary>

Use the quotient rule to differentiate $f(x) = \dfrac{x^3+5}{x^4-6}$.

| Function     | $x^3 + 5$    | $x^4 - 6$    |
|--------------|--------------|--------------|
| Derivative   | $3x^2$       | $4x^3$       |

---

### Apply the Quotient Rule:

The quotient rule states:

$$
\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}
$$

For $f(x) = \frac{x^3+5}{x^4-6}$:

$$
f'(x) = \frac{(3x^2)(x^4 - 6) - (x^3+5)(4x^3)}{(x^4 - 6)^2}
$$

---

### Detailed Simplification:

1. Expand the first term: $(3x^2)(x^4 - 6)$
   $$
   3x^2(x^4 - 6) = 3x^6 - 18x^2
   $$

2. Expand the second term: $(x^3+5)(4x^3)$
   $$
   (x^3+5)(4x^3) = 4x^6 + 20x^3
   $$

3. Combine the terms in the numerator:
   $$
   (3x^6 - 18x^2) - (4x^6 + 20x^3)
   $$

4. Distribute the negative sign:
   $$
   3x^6 - 18x^2 - 4x^6 - 20x^3
   $$

5. Combine like terms:
   $$
   (3x^6 - 4x^6) - 20x^3 - 18x^2 = -x^6 - 20x^3 - 18x^2
   $$

So the expression becomes:

$$
f'(x) = \frac{-x^6 - 20x^3 - 18x^2}{(x^4 - 6)^2}
$$

---

### Final Answer:

$$
f'(x) = \frac{-x^6 - 20x^3 - 18x^2}{(x^4 - 6)^2}
$$


:::

---

# Question 4

### Find $f'(x)$ and the equation of the line tangent to the graph of $f(x)$ at $x = 2$:

$$
f(x) = \frac{x-8}{3x-4} \quad \Rightarrow \quad f'(x) = \text{______}
$$

::: spoiler
<summary> Example: </summary>

### Given:
$$ f(x) = \frac{x-7}{4x-5} $$

We need to:
1. Find the derivative $f'(x)$ using the quotient rule.
2. Use $f'(3)$ to find the slope of the tangent line at $x = 3$.
3. Find the equation of the tangent line at $x = 3$.

---

### Step 1: Find $f'(x)$ using the quotient rule

The quotient rule is:

$$
\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}
$$

Let:
- $f(x) = x - 7$ (numerator)
- $g(x) = 4x - 5$ (denominator)

| Function     | $x-7$    | $4x-5$  |
|--------------|----------|---------|
| Derivative   | $1$      | $4$     |

Now, apply the quotient rule:

$$
f'(x) = \frac{(1)(4x - 5) - (x - 7)(4)}{(4x - 5)^2}
$$

---

### Step 2: Simplify $f'(x)$

1. Expand the first term: $(1)(4x - 5)$
   $$
   4x - 5
   $$

2. Expand the second term: $(x - 7)(4)$
   $$
   4x - 28
   $$

3. Combine the terms in the numerator:
   $$
   (4x - 5) - (4x - 28) = 4x - 5 - 4x + 28
   $$

4. Simplify:
   $$
   28 - 5 = 23
   $$

So the derivative is:

$$
f'(x) = \frac{23}{(4x - 5)^2}
$$

---

### Step 3: Find the slope at $x = 3$

To find the slope of the tangent line, evaluate $f'(x)$ at $x = 3$:

$$
f'(3) = \frac{23}{(4(3) - 5)^2} = \frac{23}{(12 - 5)^2} = \frac{23}{7^2} = \frac{23}{49}
$$

Thus, the slope of the tangent line at $x = 3$ is $\frac{23}{49}$.

---

### Step 4: Find the equation of the tangent line

The equation of a line is given by:

$$
y - y_1 = m(x - x_1)
$$

Where:
- $m$ is the slope ($m = \frac{23}{49}$).
- $(x_1, y_1)$ is the point on the curve at $x = 3$.

First, we find $f(3)$:

$$
f(3) = \frac{3 - 7}{4(3) - 5} = \frac{-4}{12 - 5} = \frac{-4}{7}
$$

So the point of tangency is $(3, -\frac{4}{7})$.

Now, plug the values into the line equation:

$$
y - \left(-\frac{4}{7}\right) = \frac{23}{49}(x - 3)
$$

Simplifying:

$$
y + \frac{4}{7} = \frac{23}{49}(x - 3)
$$

Multiply out the right-hand side:

$$
y + \frac{4}{7} = \frac{23}{49}x - \frac{69}{49}
$$

Now, subtract $\frac{4}{7}$ (or equivalently $\frac{28}{49}$) from both sides:

$$
y = \frac{23}{49}x - \frac{69}{49} - \frac{28}{49} = \frac{23}{49}x - \frac{97}{49}
$$

---

### Final Answer:

- The derivative is: 
  $$
  f'(x) = \frac{23}{(4x - 5)^2}
  $$

- The equation of the tangent line at $x = 3$ is:
  $$
  y = \frac{23}{49}x - \frac{97}{49}
  $$

![{FAAFA4E6-5166-4E18-95E2-EC075233B42A}](https://hackmd.io/_uploads/Bkp6qGIAC.png)


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