Question

find the derivative of xsinx/1+cosx
​

Answer 1

Answer:

(x cos(x) + sin(x) + sin(x) cos(x) + x) / (1 + cos(x))^2

Step-by-step explanation:

Answer 2

The derivative is $\frac{x}{2}\sec^{2} \frac{x}{2} +\tan \frac{x}{2}$ .

Given,

$\frac{x\sin x}{1+\cos x}$    ....................................................................... (1)

To find the derivative let us first simplify the given equation (1),

⇒ $\frac{x2\sin \frac{x}{2} \cos \frac{x}{2} }{1+2\cos^{2}\frac{x}{2} -1}$  ................................(from double angle formula of $\sin x$ and $\cos x$)

⇒ $\frac{2x\sin \frac{x}{2} \cos\frac{x}{2} }{2\cos^{2}\frac{x}{2} }$

⇒ $\frac{x\sin \frac{x}{2} }{\cos \frac{x}{2} }$

⇒ $x\tan\frac{x}{2}$    .................................................................. (2)

Now, on differentiating the simplified equation (2) we get,

$\frac{d }{dx}(x\tan \frac{x}{2}) = x\frac{d}{dx}(\tan \frac{x}{2})+\tan \frac{x}{2} (\frac{d}{dx}x )$

⇒ $\frac{d}{dx} (x\tan \frac{x}{2} )=x\sec^{2} \frac{x}{2} \frac{1}{2} + \tan \frac{x}{2}$

⇒ $\frac{d}{dx} (x\tan \frac{x}{2} )=\frac{x}{2}\sec^{2} \frac{x}{2} +\tan \frac{x}{2}$

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