(a) Use the method of first principles to determine the derivative of
(b) Use an appropriate method of differentiation to determine the derivative of the following functions (sim-
plify your answers as far as pnssible):
(i)
(ii)
(iii)

Ask by Carlson Wagner. in South Africa

Jan 23,2025

Question

(a) Use the method of first principles to determine the derivative of
(b) Use an appropriate method of differentiation to determine the derivative of the following functions (sim-
plify your answers as far as pnssible):
(i)
(ii)
(iii)

Ask by Carlson Wagner. in South Africa

Jan 23,2025

UpStudy AI · Accepted Answer

Certainly! Let's tackle each part of your question step by step.

---

### **Problem (a):**

**Use the method of first principles to determine the derivative of $$ f(x) = \frac{6}{x} $$.**

**Solution:**

The derivative of a function $$ f(x) $$ using first principles (the definition of the derivative) is given by:

$$
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
$$

Let's apply this to $$ f(x) = \frac{6}{x} $$:

1. **Compute $$ f(x + h) $$:**

$$
f(x + h) = \frac{6}{x + h}
$$

2. **Find the difference $$ f(x + h) - f(x) $$:**

$$
f(x + h) - f(x) = \frac{6}{x + h} - \frac{6}{x} = 6\left( \frac{1}{x + h} - \frac{1}{x} \right)
$$

3. **Simplify the difference:**

$$
6\left( \frac{1}{x + h} - \frac{1}{x} \right) = 6 \left( \frac{x - (x + h)}{x(x + h)} \right) = 6 \left( \frac{-h}{x(x + h)} \right) = \frac{-6h}{x(x + h)}
$$

4. **Divide by $$ h $$ and take the limit as $$ h \to 0 $$:**

$$
f'(x) = \lim_{h \to 0} \frac{-6h}{x(x + h)h} = \lim_{h \to 0} \frac{-6}{x(x + h)} = \frac{-6}{x^2}
$$

**Answer:**

$$
f'(x) = -\frac{6}{x^2}
$$

---

### **Problem (b)(i):**

**Differentiate $$ f(x) = \cos\left( \sqrt{\sin(\tan(\pi x))} \right) $$.**

**Solution:**

To find the derivative of a composite function, we'll use the chain rule multiple times. Let's break it down step by step.

1. **Identify the innermost function and work outward:**

- Let $$ u = \tan(\pi x) $$
   - Then $$ v = \sin(u) = \sin(\tan(\pi x)) $$
   - Next, $$ w = \sqrt{v} = \sqrt{\sin(\tan(\pi x))} $$
   - Finally, $$ f(x) = \cos(w) = \cos\left( \sqrt{\sin(\tan(\pi x))} \right) $$

2. **Differentiate each layer using the chain rule:**

- $$ \frac{du}{dx} = \pi \sec^2(\pi x) $$  
     (Derivative of $$ \tan(\pi x) $$ is $$ \pi \sec^2(\pi x) $$)
   
   - $$ \frac{dv}{du} = \cos(u) = \cos(\tan(\pi x)) $$
   
   - $$ \frac{dw}{dv} = \frac{1}{2\sqrt{v}} = \frac{1}{2\sqrt{\sin(\tan(\pi x))}} $$
   
   - $$ \frac{df}{dw} = -\sin(w) = -\sin\left( \sqrt{\sin(\tan(\pi x))} \right) $$

3. **Apply the chain rule by multiplying the derivatives:**

$$
f'(x) = \frac{df}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx}
$$

Substituting the derivatives:

$$
f'(x) = \left[ -\sin\left( \sqrt{\sin(\tan(\pi x))} \right) \right] \cdot \left[ \frac{1}{2\sqrt{\sin(\tan(\pi x))}} \right] \cdot \left[ \cos(\tan(\pi x)) \right] \cdot \left[ \pi \sec^2(\pi x) \right]
$$

4. **Simplify the expression:**

$$
f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2\sqrt{\sin(\tan(\pi x))}}
$$

**Answer:**

$$
f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}}
$$

---

### **Problem (b)(ii):**

**Differentiate $$ p(t) = \frac{\cos(t)}{1 - \sin(t)} $$.**

**Solution:**

We'll use the quotient rule for differentiation, which states:

$$
\left( \frac{N}{D} \right)' = \frac{N' D - N D'}{D^2}
$$

For $$ p(t) = \frac{\cos(t)}{1 - \sin(t)} $$:

1. **Identify $$ N(t) $$ and $$ D(t) $$:**

- $$ N(t) = \cos(t) $$
   - $$ D(t) = 1 - \sin(t) $$

2. **Find the derivatives $$ N'(t) $$ and $$ D'(t) $$:**

- $$ N'(t) = -\sin(t) $$
   - $$ D'(t) = -\cos(t) $$

3. **Apply the quotient rule:**

$$
p'(t) = \frac{(-\sin(t))(1 - \sin(t)) - \cos(t)(-\cos(t))}{(1 - \sin(t))^2}
$$

4. **Simplify the numerator:**

$$
-\sin(t)(1 - \sin(t)) + \cos^2(t) = -\sin(t) + \sin^2(t) + \cos^2(t)
$$

Recall that $$ \sin^2(t) + \cos^2(t) = 1 $$, so:

$$
-\sin(t) + 1 = 1 - \sin(t)
$$

5. **Final expression for the derivative:**

$$
p'(t) = \frac{1 - \sin(t)}{(1 - \sin(t))^2} = \frac{1}{1 - \sin(t)}
$$

**Answer:**

$$
p'(t) = \frac{1}{1 - \sin(t)}
$$

---

### **Problem (b)(iii):**

**Differentiate $$ g(x) = \ln\left( \frac{e^x}{1 + e^x} \right) $$.**

**Solution:**

First, let's simplify the expression before differentiating.

1. **Simplify $$ g(x) $$:**

$$
g(x) = \ln\left( \frac{e^x}{1 + e^x} \right) = \ln(e^x) - \ln(1 + e^x) = x - \ln(1 + e^x)
$$

2. **Differentiate $$ g(x) $$:**

$$
g'(x) = \frac{d}{dx} \left( x - \ln(1 + e^x) \right) = 1 - \frac{d}{dx} \ln(1 + e^x)
$$

3. **Differentiate $$ \ln(1 + e^x) $$:**

Using the chain rule:

$$
\frac{d}{dx} \ln(1 + e^x) = \frac{e^x}{1 + e^x}
$$

4. **Combine the results:**

$$
g'(x) = 1 - \frac{e^x}{1 + e^x} = \frac{(1 + e^x) - e^x}{1 + e^x} = \frac{1}{1 + e^x}
$$

**Answer:**

$$
g'(x) = \frac{1}{1 + e^{x}}
$$

---

### **Summary of Answers:**

- **(a)** $$ f'(x) = -\dfrac{6}{x^{2}} $$
  
- **(b)(i)**  
  $$
  f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^{2}(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}}
  $$
  
- **(b)(ii)**  
  $$
  p'(t) = \frac{1}{1 - \sin(t)}
  $$
  
- **(b)(iii)**  
  $$
  g'(x) = \frac{1}{1 + e^{x}}
  $$
**Derivatives:** - **(a)** $$ f'(x) = -\dfrac{6}{x^{2}} $$ - **(b)(i)** $$ f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^{2}(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}} $$ - **(b)(ii)** $$ p'(t) = \frac{1}{1 - \sin(t)} $$ - **(b)(iii)** $$ g'(x) = \frac{1}{1 + e^{x}} $$

(a) Use the method of first principles to determine the derivative of
(b) Use an appropriate method of differentiation to determine the derivative of the following functions (sim-
plify your answers as far as pnssible):
(i)
(ii)
(iii)

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