Let's solve **question 1** step by step as per your request: --- ## 1. \( y = \frac{1}{x} + \tan x \) ### **Step 1: Find the first derivative \( \frac{dy}{dx} \)** - The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \). - The derivative of \( \tan x \) is \( \sec^2 x \). So, \[ \frac{dy}{dx} = -\frac{1}{x^2} + \sec^2 x \] **Explanation:** We applied the power rule to \( \frac{1}{x} \) (which is \( x^{-1} \)), and the derivative of tangent is a standard result. --- ### **Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)** - The derivative of \( -\frac{1}{x^2} \) is \( 2x^{-3} = \frac{2}{x^3} \). - The derivative of \( \sec^2 x \) is \( 2\sec^2 x \tan x \) (using chain rule). So, \[ \frac{d^2y}{dx^2} = \frac{2}{x^3} + 2\sec^2 x \tan x \] **Explanation:** We differentiated each term of the first derivative separately: the power rule for \( -\frac{1}{x^2} \), and the chain rule for \( \sec^2 x \). --- ### **Final Answer** \[ \boxed{ \frac{d^2y}{dx^2} = \frac{2}{x^3} + 2\sec^2 x \tan x } \] --- Let me know if you'd like the solution for the next question!

## Solutions to Exercises 1 - 4 ### **1. \( y = \frac{1}{x} + \tan x \)** #### **Step 1: Find the first derivative \( \frac{dy}{dx} \)** \[ \frac{dy}{dx} = -\frac{1}{x^2} + \sec^2 x \] **Explanation:** We differentiate \( \frac{1}{x} \) to get \( -\frac{1}{x^2} \) and use the derivative of \( \tan x \), which is \( \sec^2 x \). --- #### **Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)** \[ \frac{d^2y}{dx^2} = \frac{2}{x^3} + 2\sec^2 x \tan x \] **Explanation:** Differentiating \( -\frac{1}{x^2} \) gives \( \frac{2}{x^3} \) and differentiating \( \sec^2 x \) results in \( 2\sec^2 x \tan x \). --- ### **Final Answer for 1** \[ \boxed{ \frac{d^2y}{dx^2} = \frac{2}{x^3} + 2\sec^2 x \tan x } \] --- ### **2. \( y = \sec x \)** #### **Step 1: Find the first derivative \( \frac{dy}{dx} \)** \[ \frac{dy}{dx} = \sec x \tan x \] **Explanation:** The derivative of \( \sec x \) is \( \sec x \tan x \). --- #### **Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)** \[ \frac{d^2y}{dx^2} = \sec x \tan^2 x + \sec^3 x \] **Explanation:** Using the product rule on \( \sec x \tan x \) yields \( \sec x \tan^2 x + \sec^3 x \). --- ### **Final Answer for 2** \[ \boxed{ \frac{d^2y}{dx^2} = \sec x \tan^2 x + \sec^3 x } \] --- ### **3. \( y = \sin^2 x \)** #### **Step 1: Find the first derivative \( \frac{dy}{dx} \)** \[ \frac{dy}{dx} = 2\sin x \cos x = \sin(2x) \] **Explanation:** Using the chain rule, \( \frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \cos x \) simplifies to \( \sin(2x) \). --- #### **Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)** \[ \frac{d^2y}{dx^2} = 2\cos(2x) \] **Explanation:** Differentiating \( \sin(2x) \) gives \( 2\cos(2x) \). --- ### **Final Answer for 3** \[ \boxed{ \frac{d^2y}{dx^2} = 2\cos(2x) } \] --- ### **4. For \( g(x) = \frac{1}{f(x)} \)** #### **Step 1: Find the inverse function \( f(x) \)** Given: \[ g(f(x)) = x \implies f(x) = \frac{1}{x} \] **Explanation:** To find the inverse, we switch \( x \) and \( y \) in the equation \( g(y) = x \). --- #### **Step 2: Find \( g'(-1) \)** Using the formula for the derivative of \( g(x) \): \[ g'(x) = -\frac{1}{(f(x))^2} \cdot f'(x) \] At \( x = -1 \): \[ g'(-1) = -\frac{1}{\left(f(-1)\right)^2} \cdot f'(-1) \] **Explanation:** We apply the chain rule to find \( g'(x) \). --- ### **Final Answer for 4** \[ g'(-1) = \text{(Calculate using specific \( f(x) \) and its derivative)} \] --- Let me know if you would like to continue with exercises 5 - 12!