# How do you differentiate f(x)=xcscx using the product rule?

Dec 3, 2015

$f ' \left(x\right) = \csc x \left(1 - x \cot x\right)$

#### Explanation:

Note: $F \left(x\right) = f \left(x\right) \cdot g \left(x\right)$
Then derivative : $F ' \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + g ' \left(x\right) \cdot f \left(x\right)$

Note: $f \left(x\right) = \csc x \Leftrightarrow f ' \left(x\right) = - \csc x \cot x$

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given $f \left(x\right) = x \csc x$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(x\right) \cdot \csc x + x \frac{d}{\mathrm{dx}} \csc x$

$= 1 \cdot \csc x + x \cdot \left(- \csc x \cot x\right)$
$= \csc x - x \csc x \cot x$
color(red)(=cscx (1- xcotx)

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FYI: How to derive derivative of $f \left(x\right) = \csc x$
$f \left(x\right) = \frac{1}{\sin x}$

(using quotient rule)

$f ' \left(x\right) = \frac{- \cos x}{{\sin}^{2} x}$

$f ' \left(x\right) = \left(\frac{1}{\sin} x\right) \cdot \left(- \cos \frac{x}{\sin} x\right)$

$f ' \left(x\right) = - \csc x \cot x$