Differentiate the following functions using the chain rule: vii) y = e^x2+3 viii) y = an(x2+2) ix) y = e^ x x) y = (x)

Let $u = x^2+3$.
Then $\frac{du}{dx} = \frac{d}{dx}(x^2+3) = 2x$.

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$.

Let $u = x^2+2$.
Then $\frac{du}{dx} = \frac{d}{dx}(x^2+2) = 2x$.

$\frac{dy}{du} = \frac{d}{du}(\tan u) = \sec^2 u$.

Let $u = \sin x$.
Then $\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$.

$\frac{dy}{du} = \frac{d}{du}(e^u) = e^u$.

Let $u = \sqrt{x} = x^{1/2}$.
Then $\frac{du}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

$\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u$.

xi) $y = \cos^2(2\sqrt{x})$

Let $u = \cos(2\sqrt{x})$.
Then $\frac{dy}{du} = \frac{d}{du}(u^2) = 2u = 2\cos(2\sqrt{x})$.

Let $v = 2\sqrt{x} = 2x^{1/2}$.
Then $\frac{dv}{dx} = \frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2}x^{-1/2} = x^{-1/2} = \frac{1}{\sqrt{x}}$.

$\frac{du}{dv} = \frac{d}{dv}(\cos v) = -\sin v = -\sin(2\sqrt{x})$.

Here $\theta = 2\sqrt{x}$.
So, $2\sin(2\sqrt{x})\cos(2\sqrt{x}) = \sin(2 \cdot 2\sqrt{x}) = \sin(4\sqrt{x})$.

xii) $y = \cos(\frac{\pi}{4} - 3x^2)$

Let $u = \frac{\pi}{4} - 3x^2$.
Then $\frac{du}{dx} = \frac{d}{dx}(\frac{\pi}{4} - 3x^2) = 0 - 3(2x) = -6x$.

$\frac{dy}{du} = \frac{d}{du}(\cos u) = -\sin u$.

xiii) $y = \ln\left(\frac{x-3}{x^2+3}\right)$

For the first term, $\frac{d}{dx}(\ln(x-3))$:
Let $u = x-3$, so $\frac{du}{dx} = 1$.
$\frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{x-3} \cdot 1 = \frac{1}{x-3}$.
For the second term, $\frac{d}{dx}(\ln(x^2+3))$:
Let $v = x^2+3$, so $\frac{dv}{dx} = 2x$.
$\frac{d}{dx}(\ln v) = \frac{1}{v} \cdot \frac{dv}{dx} = \frac{1}{x^2+3} \cdot 2x = \frac{2x}{x^2+3}$.