Using the product rule, find the derivative of each of the following functions wrt x. (i) y=(x2+2x−5)(3x2+1) (ii) y=3x4 x (iii) y=(x3−1) x

Let $u = 3x^2 + 2$ and $v = 5x - 1$.
Then $\frac{du}{dx} = 6x$ and $\frac{dv}{dx} = 5$.
The product rule states $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$.

$$f'(x) = (3x^2 + 2)(5) + (5x - 1)(6x)$$

$$f'(x) = 15x^2 + 10 + 30x^2 - 6x$$ $$f'(x) = 45x^2 - 6x + 10$$

The derivative is $\boxed{45x^2 - 6x + 10}$.

Let $u = 4x^2 - 3$ and $v = 2x + 1$.
Then $\frac{du}{dx} = 8x$ and $\frac{dv}{dx} = 2$.
The quotient rule states $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

$$\frac{dy}{dx} = \frac{(2x+1)(8x) - (4x^2-3)(2)}{(2x+1)^2}$$

$$\frac{dy}{dx} = \frac{16x^2 + 8x - 8x^2 + 6}{(2x+1)^2}$$ $$\frac{dy}{dx} = \frac{8x^2 + 8x + 6}{(2x+1)^2}$$

The derivative is $\boxed{\frac{8x^2 + 8x + 6}{(2x+1)^2}}$.

Recall that $a \log x = \log x^a$.

$$y = e^{2\log x} = e^{\log x^2}$$

$$\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x$$

The derivative is $\boxed{2x}$.