1. **Problem a:** Find $\frac{dy}{dx}$ for $y = \sqrt{x^2 + 2}$. - Rewrite $y$ as $y = (x^2 + 2)^{1/2}$. - Use the chain rule: $\frac{dy}{dx} = \frac{1}{2}(x^2 + 2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 2}}$. 2. **Problem b:** Find $f'(x)$ for $f(x) = 10^x$. - Use the exponential derivative formula: $\frac{d}{dx} a^x = a^x \ln a$. - So, $f'(x) = 10^x \ln 10$. 3. **Problem c:** Find the slope of the tangent line at $x=2$ for $f(x) = i \ln(2x) + x^2$. - Differentiate: $f'(x) = i \cdot \frac{1}{2x} \cdot 2 + 2x = \frac{i}{x} + 2x$. - Evaluate at $x=2$: $f'(2) = \frac{i}{2} + 4$. 4. **Problem d:** Find $f'(x)$ for $f(x) = \sin(\pi x) + 6x^3$. - Use chain rule and power rule: $$f'(x) = \pi \cos(\pi x) + 18x^2$$ 5. **Problem f:** Find $\frac{dy}{dx}$ for $y = \sin\left(\csc\left|x^2 + 1\right|\right)$. - Let $u = |x^2 + 1|$, since $x^2 + 1 > 0$ for all real $x$, $u = x^2 + 1$. - Then $y = \sin(\csc(u))$. - Differentiate stepwise: $$\frac{dy}{dx} = \cos(\csc(u)) \cdot (-\csc(u) \cot(u)) \cdot \frac{du}{dx}$$ - Since $\frac{du}{dx} = 2x$, the derivative is: $$\frac{dy}{dx} = -2x \cos(\csc(x^2 + 1)) \csc(x^2 + 1) \cot(x^2 + 1)$$ **Final answers:** - a) $\frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 2}}$ - b) $f'(x) = 10^x \ln 10$ - c) Slope at $x=2$ is $\frac{i}{2} + 4$ - d) $f'(x) = \pi \cos(\pi x) + 18x^2$ - f) $\frac{dy}{dx} = -2x \cos(\csc(x^2 + 1)) \csc(x^2 + 1) \cot(x^2 + 1)$