Find the derivatives of the following functions: y = ex2x3 + 1 and y = sqrt(4x + 1)x2 − 5

Let $u = e^{x^2}$ and $v = x^3 + 1$.

Using the chain rule for $u$: $u' = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot (2x) = 2xe^{x^2}$.
For $v$: $v' = 3x^2$.

3. $y = \frac{\sqrt{4x + 1}}{x^2 - 5}$

Let $u = \sqrt{4x + 1} = (4x + 1)^{1/2}$ and $v = x^2 - 5$.

Using the chain rule for $u$: $u' = \frac{1}{2}(4x + 1)^{-1/2} \cdot \frac{d}{dx}(4x + 1) = \frac{1}{2}(4x + 1)^{-1/2} \cdot 4 = 2(4x + 1)^{-1/2} = \frac{2}{\sqrt{4x + 1}}$.
For $v$: $v' = 2x$.

Multiply the numerator by $\sqrt{4x+1}$ to clear the fraction in the numerator.

5. $f(x) = \frac{\cos(x^2)}{x}$

Let $u = \cos(x^2)$ and $v = x$.

Using the chain rule for $u$: $u' = -\sin(x^2) \cdot \frac{d}{dx}(x^2) = -\sin(x^2) \cdot (2x) = -2x\sin(x^2)$.
For $v$: $v' = 1$.

7. $y = \frac{(x+1)^2}{(x-1)^2}$

Let $u = (x+1)^2$ and $v = (x-1)^2$.

Using the chain rule for $u$: $u' = 2(x+1) \cdot \frac{d}{dx}(x+1) = 2(x+1) \cdot 1 = 2(x+1)$.
Using the chain rule for $v$: $v' = 2(x-1) \cdot \frac{d}{dx}(x-1) = 2(x-1) \cdot 1 = 2(x-1)$.

9. $y = \frac{1 + \sin x}{1 + \cos x}$

Let $u = 1 + \sin x$ and $v = 1 + \cos x$.

For $u$: $u' = \cos x$.
For $v$: $v' = -\sin x$.

Recall the identity $\sin^2 x + \cos^2 x = 1$.