---
# System prepended metadata

title: In-Class Activity 3.4
tags: ['3.4']

---

[toc]

## Question 1
Decompose each function $C(x)$ into an outside function $f(u)$ and an inside function $u=g(x)$ such that $C(x)=(f \circ g)(x)=f(g(x))$.

### (a)
$$C(x) =  2+4x^2-4x^8$$

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<summary> Example: </summary>




$C(x)=3+5x^3-4x^6$

$f(u)=3+5u-4u^2$

$u=g(x)=x^3$


Can put it in a nice little table:

| Outside | Inside |
|----   | ---- |
|$f(u)=3+5u-4u^2$ | $u=g(x)=x^3$ |


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---

### (b)
$$C(x) = \frac{1-x^2}{2+3x^4}$$

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<summary> Example: </summary>

$C(x)=\dfrac{3-x^4}{4+5x^8}$

$f(u)=\dfrac{3-u}{4+5u^2}$

$u=g(x)=x^4$


| Outside | Inside |
|----   | ---- |
|$f(u)=\dfrac{3-u}{4+5u^2}$ | $u=g(x)=x^4$ |

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---

### (c)
$$C(x) = \sqrt{1+x^2}$$

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<summary> Example: </summary>


$C(x)=\sqrt{4+5x^3}$

$f(u)=\sqrt{u}=u^{1/2}$

$u=g(x)=4+5x^3$

| Outside          | Inside        |
|------------------|---------------|
| $f(u) = \sqrt{u} = u^{1/2}$ | $u=g(x) = 4 + 5x^3$ |


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---

### (d)
$$C(x) = \frac{1}{3x^2-2x}$$

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<summary> Example: </summary>

$C(x)=\dfrac{2}{4x^3-5x}$

$f(u)=\dfrac{2}{u}$

$u=g(x)=4x^3-5x$

| Outside          | Inside        |
|------------------|---------------|
| $f(u) = \dfrac{2}{u}$ | $u=g(x) = 4x^3 - 5x$ |


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### (e)
$$C(x) = \frac{1+2e^x}{e^{2x}-1}$$

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<summary> Example: </summary>

$C(x)=\dfrac{2+3e^{x}}{e^{3x}-1}$

$f(u)=\dfrac{2+3u}{u^3-1}$

$u=g(x)=e^x$

| Outside          | Inside        |
|------------------|---------------|
| $f(u) = \dfrac{2+3u}{u^3-1}$ | $u=g(x) = e^x$ |


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### (f)
$$C(x) = e^{2x^3-5}$$

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<summary> Example: </summary>

$C(x)=e^{5x^2-4}$

$f(u)=e^u$

$g(x)=5x^2-4$

| Outside          | Inside        |
|------------------|---------------|
| $f(u) = e^u$     | $u=g(x) = 5x^2 - 4$ |


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### (g)
$$C(x) = \ln(2x^3-x^2)$$

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<summary> Example: </summary>

$C(x)=\ln(3x^4+2x^3)$

$f(u)=\ln u$

$g(x)=3x^4+2x^3$

| Outside          | Inside        |
|------------------|---------------|
| $f(u) = \ln u$   | $u=g(x) = 3x^4 + 2x^3$ |


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---

## Question 2
Replace ? with an expression that will make the indicated equation valid.

### (a)
$$
   \frac{d}{dx}(4-2x^2)^3 = 3(4-2x^2)^2\: ? 
$$

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<summary> Example: </summary>

$$
   \frac{d}{dx}(5+7x^3)^4 = 4(5+7x^3)^3?
$$

- The inside function is $u=5+7x^3$.
- The derivative of the inside function is $u'=21x^2$

Thus 
$$
   \frac{d}{dx}(5+7x^3)^4 = 4(5+7x^3)^3(21x^2)
$$

This can be shown in this neat table:

|   | Outside          | Inside        |
| --- |------------------|---------------|
| Original | $f(u) =u^4$   | $u=g(x) = 5+7x^3$ |
| Derivative | $f'(u)=4u^3$ | $u'=g'(x)=21x^2$ |


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---

### (b)
$$
   \frac{d}{dx}(3x^2+7)^5 = 5(3x^2+7)^4\: ? 
$$

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<summary> Example: </summary>

$$
   \frac{d}{dx}(4x^5+6)^7 = 7(4x^5+6)^6 ?
$$

- The inside function is $u = 4x^5+6$.
- The derivative of the inside function is $u' = 20x^4$.

Thus, 
$$
   \frac{d}{dx}(4x^5+6)^7 = 7(4x^5+6)^6 (20x^4)
$$

|  | Outside          | Inside        |
|---|------------------|---------------|
| Original | $f(u) =u^7$   | $u=g(x) = 4x^5+6$ |
| Derivative | $f'(u)=7u^6$ | $u'=g'(x)=20x^4$ |

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---

### (c)
$$
   \frac{d}{dx}e^{x^2+1} = e^{x^2+1}\: ? 
$$

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<summary> Example: </summary>

$$
   \frac{d}{dx}e^{x^4+2} = e^{x^4+2} ?
$$

- The inside function is $u = x^4 + 2$.
- The derivative of the inside function is $u' = 4x^3$.

Thus, 
$$
   \frac{d}{dx}e^{x^4+2} = e^{x^4+2} (4x^3)
$$

|  | Outside          | Inside        |
|---|------------------|---------------|
| Original | $f(u) = e^u$   | $u = x^4 + 2$ |
| Derivative | $f'(u) = e^u$ | $u' = 4x^3$ |


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---

### (d)
$$
   \frac{d}{dx}e^{4x-2} = e^{4x-2}\: ? 
$$

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<summary> Example: </summary>

$$
   \frac{d}{dx}e^{3x-5} = e^{3x-5} ?
$$

- The inside function is $u = 3x - 5$.
- The derivative of the inside function is $u' = 3$.

Thus, 
$$
   \frac{d}{dx}e^{3x-5} = e^{3x-5} (3)
$$

|  | Outside          | Inside        |
|---|------------------|---------------|
| Original | $f(u) = e^u$   | $u = 3x - 5$ |
| Derivative | $f'(u) = e^u$ | $u' = 3$ |


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---

## Question 3
Find $f'(x)$ and simplify:

### (a)
$$
   f(x) = (3x^2+5)^5
$$

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<summary> Example: </summary>

$$
   f(x) = (4x^3+5)^7
$$

- The outside function is $u^7$.
- The derivative of the outside function is $7u^6$.
- The inside function is $4x^3+5$
- The derivative of the inside function is $12x^2$.


According to the Chain Rule, the derivative is $\text{Out}' \cdot \text{In}'=7u^6 \cdot 12x^2$

$$
f'(x)=7(4x^3+5)^6(12x^2)
$$

|  | Outside          | Inside        |
|---|------------------|---------------|
| Original  | $f(u) = u^7$    | $u = 4x^3 + 5$ |
| Derivative | $f'(u) = 7u^6$ | $u' = 12x^2$ |


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---

### (b)
$$
   f(x) = 2\ln(x^2-3x+4)
$$

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<summary> Example: </summary>





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

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

Simplifying it further:

$$f'(x)=\dfrac{6x-15}{x^2-5x+4}$$

|  | Outside          | Inside        |
|---|------------------|---------------|
| Original  | $f(u) = 3\ln(u)$    | $u = x^2 - 5x + 4$ |
| Derivative | $f'(u) = \dfrac{3}{u}$ | $u' = 2x - 5$ |


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---

## Question 4
Find $f'(x)$ and the equation of the line tangent to the graph of $f(x)$ at $x=e$.

$$f(x) = \sqrt{\ln(x)}$$

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<summary> Example: </summary>

$$f(x)=\sqrt[3]{\ln x}$$

Find the equation of the tangent line to the graph of $f(x)$ at $x=e$.

### Problem
$$ f(x) = \sqrt[3]{\ln x} $$

Find the equation of the tangent line to the graph of $f(x)$ at $x = e$.

### Solution

1. **Find the derivative $f'(x)$:**

We need to apply the chain rule. First, rewrite $f(x)$ as:

$$
f(x) = (\ln x)^{1/3}
$$

Now, differentiate $f(x)$:

$$
f'(x) = \frac{1}{3}(\ln x)^{-2/3} \cdot \frac{1}{x}
$$

So the derivative is:

$$
f'(x) = \frac{1}{3x(\ln x)^{2/3}}
$$


|  | Outside          | Inside        |
|---|------------------|---------------|
| Original  | $f(u) = u^{1/3}$    | $u = \ln x$ |
| Derivative | $f'(u) = \dfrac{1}{3}u^{-2/3}$ | $u' = \dfrac{1}{x}$ |


2. **Evaluate the derivative at $x = e$:**

Substitute $x = e$ into the derivative:

$$
f'(e) = \frac{1}{3e(\ln e)^{2/3}}
$$

Since $\ln e = 1$, we have:

$$
f'(e) = \frac{1}{3e}
$$

3. **Find the point $(e, f(e))$:**

Now, evaluate $f(e)$:

$$
f(e) = \sqrt[3]{\ln e} = \sqrt[3]{1} = 1
$$

So the point of tangency is $(e, 1)$.

4. **Find the equation of the tangent line:**

The equation of the tangent line is given by the point-slope form:

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

Here, $m = f'(e) = \frac{1}{3e}$ and $(x_1, y_1) = (e, 1)$. Substituting these values:

$$
y - 1 = \frac{1}{3e}(x - e)
$$

Thus, the equation of the tangent line is:

$$
y = \frac{1}{3e}(x - e) + 1
$$


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