Differentiate the following functions: I. y = 1/x + sqrt(x) II. y = x III. y = 2x x2

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Here are the differentiations for Question 5:

I. Differentiate $y = \frac{1}{x} + \sqrt{x}$

Rewrite the function using exponents.
$y = x^{-1} + x^{1/2}$
Apply the power rule for differentiation, $\frac{d}{dx}(x^n) = nx^{n-1}$ .
$\frac{dy}{dx} = -1 \cdot x^{-1-1} + \frac{1}{2} \cdot x^{1/2-1}$ $\frac{dy}{dx} = -x^{-2} + \frac{1}{2}x^{-1/2}$
Rewrite with positive exponents and radicals.
$\frac{dy}{dx} = \boxed{-\frac{1}{x^2} + \frac{1}{2\sqrt{x}}}$
II. Differentiate $y = \csc x$
Recall the standard derivative of $\csc x$ .
$\frac{dy}{dx} = \boxed{-\csc x \cot x}$
III. Differentiate $y = 2x \cos x^2$
Use the product rule $\frac{d}{dx}(uv) = u'v + uv'$ , where $u = 2x$ and $v = \cos x^2$ .
Find the derivatives of $u$ and $v$ .
$u = 2x \implies u' = 2$
For $v = \cos x^2$ , use the chain rule. Let $w = x^2$ , so $v = \cos w$ .
$\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx} = (-\sin w) \cdot (2x) = -2x \sin x^2$
So, $v' = -2x \sin x^2$ .
Apply the product rule.
$\frac{dy}{dx} = (2)(\cos x^2) + (2x)(-2x \sin x^2)$ $\frac{dy}{dx} = 2 \cos x^2 - 4x^2 \sin x^2$ $\frac{dy}{dx} = \boxed{2 \cos x^2 - 4x^2 \sin x^2}$

Differentiate the following functions: I. y = (1)/(x) + sqrt(x) II. y = x III. y = 2x x2