# How do you differentiate f(x)=(e^x-3x)(cotx+x^2) using the product rule?

Mar 1, 2016

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

#### Explanation:

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

Applying product rule as : ${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = {e}^{x} - 3 x , g = \setminus \cot \left(x\right) + {x}^{2}$

$= \setminus \frac{d}{\mathrm{dx}} \left({e}^{x} - 3 x\right) \left(\setminus \cot \left(x\right) + {x}^{2}\right) + \setminus \frac{d}{\mathrm{dx}} \left(\setminus \cot \setminus \left(x\right) + {x}^{2}\right) \left({e}^{x} - 3 x\right)$............ $e {q}^{n} \left(i\right)$

Here,
$\setminus \frac{d}{\mathrm{dx}} \left({e}^{x} - 3 x\right) = {e}^{x} - 3$
{Applying sum and differnce rule as: ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

$= \setminus \frac{d}{\mathrm{dx}} \left({e}^{x}\right) - \setminus \frac{d}{\mathrm{dx}} \left(3 x\right)$ and

$\setminus \frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$, $\setminus \frac{d}{\mathrm{dx}} \left(3 x\right) = 3$}

$= {e}^{x} - 3$............. $e {q}^{n} \left(i i\right)$

Again,

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

{Applying sum and differnce rule as: ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$
$= \setminus \frac{d}{\mathrm{dx}} \left(\setminus \cot \left(x\right)\right) + \setminus \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$ and

and we know the common derivative,

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \cot \left(x\right)\right) = - \setminus \frac{1}{\setminus {\sin}^{2} \left(x\right)}$ and also we know,\frac{d}

{dx}(x^2)=2x}

$= - \setminus \frac{1}{\setminus {\sin}^{2} \left(x\right)} + 2 x$ ............. $e {q}^{n} \left(i i i\right)$

Finally,from $e {q}^{n} \left(i\right) , \left(i i\right) \mathmr{and} \left(i i i\right)$, we get,

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