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To solve the given math problems, we can use differentiation rules and properties of logarithmic, exponential, and trigonometric functions. The derivative of a function y with respect to x, denoted as $\frac{dy}{dx}$, represents the rate of change of y with respect to x. i) For the function $y=2^x$, we use the property that $\frac{d}{dx}(a^x) = a^x \cdot \ln(a)$: $\frac{dy}{dx} = 2^x \cdot \ln(2)$ h) For the function $y=\sqrt{2x-1}$, we use the power rule for differentiation: $\frac{dy}{dx} = \frac{1}{2\sqrt{2x-1}}$ j) For the function $y=3^{x^2+1}$, we apply the chain rule and the property that $\frac{d}{dx}(a^{f(x)}) = a^{f(x)} \cdot f'(x) \cdot \ln(a)$: $\frac{dy}{dx} = 3^{x^2+1} \cdot (2x) \cdot \ln(3)$ k) For the function $y=\ln(1+2x)$, we use the derivative of natural logarithm: $\frac{dy}{dx} = \frac{1}{1+2x} \cdot 2$ Simplify to get: $\frac{dy}{dx} = \frac{2}{1+2x}$ l) For the function $y=\log_2(x^3+2x)$, we use the chain rule and derivative of logarithmic functions: $\frac{dy}{dx} = \frac{3x^2 + 2}{(x^3 + 2x)\ln(2)}$ m) For the function $y=e^{1-3x}$, we use the property that $\frac{d}{dx}(e^{f(x)}) = e^{f(x)} \cdot f'(x)$: $\frac{dy}{dx} = -3e^{1-3x}$ n) For the function $y=\tan(x-\frac{5\pi}{2})$, we use trigonometric identities and derivatives: Using $\tan'(u)=\sec(u)^2$ where u is a differentiable function of x, we have: $\tan'(u)=\sec(u)^22$ So, $\tan'(u)=tan(x-\pi/4)^22$ Therefore, $\tan'(u)=tan(x-\pi/4)^22$ These are all solutions to find $\frac {dy} {dx}$ for each given function.