How do you differentiate f(x)=cos5x * cot3x using the product rule?

Jul 16, 2016

$- 5 \sin 5 x \cot 3 x - 3 {\csc}^{2} \left(3 x\right) \cos 5 x$

Explanation:

The derivative of a product is stated as follows:
$\textcolor{b l u e}{\left(u \left(x\right) \cdot v \left(x\right)\right) ' = u ' \left(x\right) \cdot v \left(x\right) + v ' \left(x\right) \cdot u \left(x\right)}$

Take $u \left(x\right) = \cos \left(5 x\right)$ and $v \left(x\right) = \cot \left(3 x\right)$
Let's find $u ' \left(x\right)$ and $v ' \left(x\right)$

Knowing the derivative of trigonometric function that says:
$\left(\cos y\right) ' = - y ' \sin y$ and
$\left(\cot \left(y\right)\right) ' = - y ' \left({\csc}^{2} y\right)$
So,
$u ' \left(x\right) = \left(\cos 5 x\right) ' = - \left(5 x\right) ' \sin 5 x = - 5 \sin 5 x$

$v ' \left(x\right) = \left(\cot 3 x\right) ' = - \left(3 x\right) ' {\csc}^{2} \left(3 x\right) = - 3 {\csc}^{2} \left(3 x\right)$

Thus,
$\textcolor{b l u e}{f ' \left(x\right) = \left(u \left(x\right) \cdot v \left(x\right)\right) '}$

Substituting $u ' \left(x\right)$ and $v ' \left(x\right)$ in the above property we have:

$= - 5 \sin 5 x \cot 3 x - 3 {\csc}^{2} \left(3 x\right) \cos 5 x$