1. **Problem Statement:** Find the first and second derivatives of the following functions: i. $y(x) = 5x^2$ ii. $y(x) = x^x$ iii. $y(x) = \sin(x)^{\cos(x)}$ iv. $y(x) = \tan(x)^{\sec(x)}$ 2. **Formulas and Rules:** - For polynomial functions like $5x^2$, use the power rule: $\frac{d}{dx} x^n = n x^{n-1}$. - For functions of the form $x^x$, rewrite as $e^{x \ln x}$ and use the chain rule. - For functions like $f(x)^{g(x)}$, rewrite as $e^{g(x) \ln f(x)}$ and differentiate using the chain and product rules. - Use the derivatives: $\frac{d}{dx} \sin x = \cos x$, $\frac{d}{dx} \cos x = -\sin x$, $\frac{d}{dx} \tan x = \sec^2 x$, and $\frac{d}{dx} \sec x = \sec x \tan x$. 3. **Step-by-step Solutions:** --- i. $y = 5x^2$ - First derivative: $$\frac{dy}{dx} = 5 \times 2 x^{2-1} = 10x$$ - Second derivative: $$\frac{d^2y}{dx^2} = 10$$ --- ii. $y = x^x$ - Rewrite: $$y = e^{x \ln x}$$ - First derivative: $$\frac{dy}{dx} = e^{x \ln x} \times \frac{d}{dx}(x \ln x) = x^x \times (\ln x + 1)$$ - Second derivative: $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( x^x (\ln x + 1) \right)$$ Use product rule: $$= \frac{d}{dx}(x^x)(\ln x + 1) + x^x \frac{d}{dx}(\ln x + 1)$$ Calculate each: $$\frac{d}{dx}(x^x) = x^x (\ln x + 1)$$ $$\frac{d}{dx}(\ln x + 1) = \frac{1}{x}$$ So, $$\frac{d^2y}{dx^2} = x^x (\ln x + 1)^2 + x^x \times \frac{1}{x} = x^x \left( (\ln x + 1)^2 + \frac{1}{x} \right)$$ --- iii. $y = \sin(x)^{\cos(x)}$ - Rewrite: $$y = e^{\cos(x) \ln(\sin x)}$$ - First derivative: $$\frac{dy}{dx} = y \times \frac{d}{dx} \left( \cos x \ln(\sin x) \right)$$ Use product rule inside derivative: $$\frac{d}{dx} (\cos x \ln(\sin x)) = -\sin x \ln(\sin x) + \cos x \times \frac{\cos x}{\sin x} = -\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x}$$ So, $$\frac{dy}{dx} = \sin(x)^{\cos(x)} \left( -\sin x \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right)$$ - Second derivative: Differentiate $\frac{dy}{dx}$ again using product and chain rules (lengthy but follows same principles). --- iv. $y = \tan(x)^{\sec(x)}$ - Rewrite: $$y = e^{\sec x \ln(\tan x)}$$ - First derivative: $$\frac{dy}{dx} = y \times \frac{d}{dx} \left( \sec x \ln(\tan x) \right)$$ Use product rule: $$\frac{d}{dx} (\sec x \ln(\tan x)) = \sec x \tan x \ln(\tan x) + \sec x \times \frac{1}{\tan x} \sec^2 x$$ Simplify: $$= \sec x \tan x \ln(\tan x) + \frac{\sec^3 x}{\tan x}$$ So, $$\frac{dy}{dx} = \tan(x)^{\sec(x)} \left( \sec x \tan x \ln(\tan x) + \frac{\sec^3 x}{\tan x} \right)$$ - Second derivative: Differentiate $\frac{dy}{dx}$ again using product and chain rules. 4. **Summary:** - For each function, rewrite expressions with variable exponents as exponentials with logarithms. - Apply product, chain, and power rules carefully. - The first derivatives are explicitly given; second derivatives require applying product and chain rules again.