Question

Differentiate w.r.t.x :
$y = \frac{ {x}^{2}sinx }{x + cosx}$
​

Answer 1

Hey bud,

We have,

$y = \frac{ {x}^{2} \sin \: x }{x + \cos \: x}$

Using Lebnitz method for differentiating,

Where
$d( \frac{u}{v} ) = \frac{vdu - udv}{ {v}^{2} }$

substituing the values we have,

$\frac{ (x + \cos \: x)(d \sin \: x ) - \sin(x) (d(x + \cos \: x))}{ {(x + \cos(x)) }^{2} } \\ \\ = \frac{ \cos(x) (x + \cos(x) ) - \sin(x) (1 - \sin(x)) }{ {(x + \cos(x)) }^{2} } \\ \\ = \frac{x \cos(x) + { \cos }^{2}x - \sin \: x + { \sin }^{2} x }{ {(x + \cos(x)) }^{2} } \\ \\ = \frac{x \cos(x) - \sin(x) }{ {(x + { \cos}^{2} x)}^{2} }$

Hence huh!