How do you differentiate f(x) = x^3/(xcotx+1) using the quotient rule?

Jun 21, 2016

$\frac{d}{\mathrm{dx}} \left(\frac{{x}^{3}}{x \cot \left(x\right) + 1}\right) = \frac{2 {x}^{3} {\sin}^{2} \left(x\right) \cot \left(x\right) + 3 {x}^{2} {\sin}^{2} \left(x\right) + {x}^{4}}{{\sin}^{2} \left(x\right) {\left(x \cot \left(x\right) + 1\right)}^{2}}$

Explanation:

$\frac{d}{\mathrm{dx}} \left(\frac{{x}^{3}}{x \cot \left(x\right) + 1}\right)$

Applying quotient rule,

${\left(\frac{f}{g}\right)}^{'} = \frac{{f}^{'} \cdot g - {g}^{'} \cdot f}{{g}^{2}}$

$= \frac{\frac{d}{\mathrm{dx}} \left({x}^{3}\right) \left(x \cot \left(x\right) + 1\right) - \frac{d}{\mathrm{dx}} \left(x \cot \left(x\right) + 1\right) {x}^{3}}{{\left(x \cot \left(x\right) + 1\right)}^{2}}$

We know,
$\frac{d}{\mathrm{dx}} \left({x}^{3}\right) = 3 {x}^{2}$
$\frac{d}{\mathrm{dx}} \cot \left(x\right) + 1 = - \frac{x}{{\sin}^{2} \left(x\right)} + \cot \left(x\right)$

so,$= \frac{3 {x}^{2} \left(x \cot \left(x\right) + 1\right) - \left(- \frac{x}{{\sin}^{2} \left(x\right)} + \cot \left(x\right)\right) {x}^{3}}{{\left(x \cot \left(x\right) + 1\right)}^{2}}$

Simplifying it,
$\frac{2 {x}^{3} {\sin}^{2} \left(x\right) \cot \left(x\right) + 3 {x}^{2} {\sin}^{2} \left(x\right) + {x}^{4}}{{\sin}^{2} \left(x\right) {\left(x \cot \left(x\right) + 1\right)}^{2}}$