Question

Find the derivative of sinx+cosx/sinx-cosx

Answer 1

SOLUTION

TO DETERMINE

The derivative of

$\displaystyle \sf{ \frac{ \sin x + \cos x}{ \sin x - \cos x} }$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \frac{d}{dx} \bigg( \frac{u}{v} \bigg) = \frac{ \displaystyle \sf{v \frac{du}{dx} - u \frac{dv}{dx}} }{ {v}^{2} } }$

EVALUATION

$\displaystyle \sf{ \frac{d}{dx} \bigg( \frac{ \sin x + \cos x}{ \sin x - \cos x} \bigg)}$

$\displaystyle \sf{ = \frac{(\sin x - \cos x) \frac{d}{dx} ( \sin x + \cos x) -(\sin x + \cos x) \frac{d}{dx} ( \sin x - \cos x) }{{ (\sin x - \cos x)}^{2} } }$

$\displaystyle \sf{ = \frac{(\sin x - \cos x) ( \cos x - \sin x) -(\sin x + \cos x) ( \cos x + \sin x) }{{ (\sin x - \cos x)}^{2} } }$

$\displaystyle \sf{ = \frac{ { - ( \sin x - \cos x)}^{2} - {( \cos x + \sin x)}^{2} }{{ (\sin x - \cos x)}^{2} } }$

$\displaystyle \sf{ = \frac{ - \bigg[ { ( \sin x - \cos x)}^{2} + {( \cos x + \sin x)}^{2} \bigg] }{{ (\sin x - \cos x)}^{2} } }$

$\displaystyle \sf{ = \frac{ - 2 { ( {\sin}^{2} x + { \cos}^{2} x)} }{{ (\sin x - \cos x)}^{2} } }$

$\displaystyle \sf{ = \frac{ - 2 }{{ (\sin x - \cos x)}^{2} } }$

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