How do you differentiate given y = 2x (x^(1/2) - cot x)?

Jun 3, 2016

$= \frac{2 x + 3 \sqrt{x} {\sin}^{2} \left(x\right) - 2 {\sin}^{2} \left(x\right) \cot \left(x\right)}{{\sin}^{2} \left(x\right)}$

Explanation:

$\frac{d}{\mathrm{dx}} \left(2 x \left(\sqrt{x} - \cot \left(x\right)\right)\right)$

Taking the constant out,
${\left(a \setminus \cdot f\right)}^{'} = a \setminus \cdot {f}^{'}$

$= 2 \frac{d}{\mathrm{dx}} \left(x \left(\sqrt{x} - \cot \left(x\right)\right)\right)$

Applying the product rule,
${\left(f \cdot g\right)}^{'} = {f}^{'} \cdot g + f \cdot {g}^{'}$
$f = x : g = \sqrt{x} - \cot x$

$= 2 \left(\frac{d}{\mathrm{dx}} \left(x\right) \left(\sqrt{x} - \cot \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(\sqrt{x} - \cot \left(x\right)\right) x\right)$

We know,
$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

$\frac{d}{\mathrm{dx}} \left(\sqrt{x} - \cot \left(x\right)\right) = \frac{1}{2 \sqrt{x}} + \frac{1}{{\sin}^{2} \left(x\right)}$

$= 2 \left(1 \left(\sqrt{x} - \setminus \cot \left(x\right)\right) + \left(\frac{1}{2 \sqrt{x}} + \frac{1}{{\sin}^{2} \left(x\right)}\right) x\right)$

Simplify,
$= \frac{2 x + 3 \sqrt{x} {\sin}^{2} \left(x\right) - 2 {\sin}^{2} \left(x\right) \cot \left(x\right)}{{\sin}^{2} \left(x\right)}$