UUESTION 4 Use an appropriate method off differentiation to determine the derivative of the following plify your answers as far as possible): (i) $$ f(x)=\sqrt{x}+3^{x}-\frac{2}{x^{3}} $$ (ii) $$ g(x)=(\sin x)(\ln x) $$ (cos $$ x) $$ (iii) $$ h(x)=\left(x^{2}-2\right)^{x+1} $$

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Sure, let's find the derivatives of each function step by step.

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### **Problem (i)**

$$ f(x) = \sqrt{x} + 3^{x} - \frac{2}{x^{3}} $$

**Solution:**

We'll differentiate each term separately.

1. **Derivative of $$\sqrt{x}$$:**
   
   $$
   \sqrt{x} = x^{1/2} \
   \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}
   $$

2. **Derivative of $$3^{x}$$:**
   
   The derivative of $$a^{x}$$ with respect to $$x$$ is $$a^{x} \ln a$$.
   
   $$
   \frac{d}{dx} 3^{x} = 3^{x} \ln 3
   $$

3. **Derivative of $$-\frac{2}{x^{3}}$$:**
   
   Rewrite the term as $$-2x^{-3}$$ and differentiate.
   
   $$
   \frac{d}{dx} (-2x^{-3}) = -2 \cdot (-3) x^{-4} = 6x^{-4} = \frac{6}{x^{4}}
   $$

**Combine the derivatives:**

$$
f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^{4}}
$$

---

### **Problem (ii)**

$$ g(x) = (\sin x)(\ln x) (\cos x) $$

**Solution:**

First, it's easier to handle this by recognizing it as a product of three functions:
$$
g(x) = \sin x \cdot \ln x \cdot \cos x
$$

To differentiate $$g(x)$$, we'll use the **product rule** for multiple functions. The product rule for three functions states:
$$
\frac{d}{dx} [u \cdot v \cdot w] = u' \cdot v \cdot w + u \cdot v' \cdot w + u \cdot v \cdot w'
$$

Let:
$$
u = \sin x \
v = \ln x \
w = \cos x
$$

Compute each derivative:

1. **$$u' = \cos x$$**

2. **$$v' = \frac{1}{x}$$**

3. **$$w' = -\sin x$$**

Apply the product rule:

$$
g'(x) = (\cos x)(\ln x)(\cos x) + (\sin x)\left(\frac{1}{x}\right)(\cos x) + (\sin x)(\ln x)(-\sin x)
$$

Simplify each term:

$$
g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x
$$

We can factor $$\ln x$$ from the first and third terms:

$$
g'(x) = (\cos^2 x - \sin^2 x) \cdot \ln x + \frac{\sin x \cos x}{x}
$$

Alternatively, you can leave it in the expanded form as:

$$
g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x
$$

---

### **Problem (iii)**

$$ h(x) = \left( x^{2} - 2 \right)^{x + 1} $$

**Solution:**

This function has a variable in both the base and the exponent. To differentiate such functions, we'll use **logarithmic differentiation**.

1. **Take the natural logarithm of both sides:**

$$
   \ln h(x) = (x + 1) \cdot \ln (x^{2} - 2)
   $$

2. **Differentiate both sides with respect to $$x$$:**

Using the chain rule on the left and the product rule on the right.

$$
   \frac{h'(x)}{h(x)} = \frac{d}{dx} \left[ (x + 1) \cdot \ln (x^{2} - 2) \right]
   $$

Apply the product rule:

$$
   \frac{h'(x)}{h(x)} = \frac{d}{dx}(x + 1) \cdot \ln (x^{2} - 2) + (x + 1) \cdot \frac{d}{dx} \ln (x^{2} - 2)
   $$

Compute each derivative:

- $$\frac{d}{dx}(x + 1) = 1$$
   - $$\frac{d}{dx} \ln (x^{2} - 2) = \frac{2x}{x^{2} - 2}$$ (using the chain rule)

Substitute back:

$$
   \frac{h'(x)}{h(x)} = 1 \cdot \ln (x^{2} - 2) + (x + 1) \cdot \frac{2x}{x^{2} - 2}
   $$

3. **Solve for $$h'(x)$$:**

Multiply both sides by $$h(x)$$:

$$
   h'(x) = h(x) \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right]
   $$

4. **Substitute back $$h(x)$$:**

$$
   h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right]
   $$

**Simplified Form:**

$$
h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right]
$$

---

**Summary of Derivatives:**

1. **$$f'(x)$$:**
   $$
   f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^{4}}
   $$

2. **$$g'(x)$$:**
   $$
   g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x
   $$
   *Or alternatively:*
   $$
   g'(x) = (\cos^2 x - \sin^2 x) \cdot \ln x + \frac{\sin x \cos x}{x}
   $$

3. **$$h'(x)$$:**
   $$
   h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right]
   $$

If you have any further questions or need additional clarification on any of the steps, feel free to ask!
**Derivatives of the Functions:** 1. **For $$ f(x) = \sqrt{x} + 3^{x} - \frac{2}{x^{3}} $$:** $$ f'(x) = \frac{1}{2\sqrt{x}} + 3^{x} \ln 3 + \frac{6}{x^{4}} $$ 2. **For $$ g(x) = (\sin x)(\ln x)(\cos x) $$:** $$ g'(x) = \cos^2 x \cdot \ln x + \frac{\sin x \cos x}{x} - \sin^2 x \cdot \ln x $$ *Alternatively:* $$ g'(x) = (\cos^2 x - \sin^2 x) \cdot \ln x + \frac{\sin x \cos x}{x} $$ 3. **For $$ h(x) = \left( x^{2} - 2 \right)^{x + 1} $$:** $$ h'(x) = \left( x^{2} - 2 \right)^{x + 1} \left[ \ln (x^{2} - 2) + \frac{2x(x + 1)}{x^{2} - 2} \right] $$ If you have any questions about these derivatives, feel free to ask!

UUESTION 4 Use an appropriate method off differentiation to determine the derivative of the following plify your answers as far as possible): (i) \( f(x)=\sqrt{x}+3^{x}-\frac{2}{x^{3}} \) (ii) \( g(x)=(\sin x)(\ln x) \) (cos \( x) \) (iii) \( h(x)=\left(x^{2}-2\right)^{x+1} \)

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