# Find the derivative of y=3tan^-1(x+sqrt(1+x^2))?

May 16, 2018

(dy)/(dx)=3/(2(1+x^2)

#### Explanation:

We know that.

color(violet)((1)cos(pi/2-theta)=sintheta and sin(pi/2-theta)=costheta

color(blue)((2) (1+cosalpha)=2cos^2(alpha/2) and sinalpha=2sin(alpha/2)cos(alpha/2)

color(brown)((3)cot(pi/2-alpha)=tanalpha

color(red)((4)tan^-1(tan alpha)=alpha ,where, alpha in(-pi/2,pi/2)

Here,

$y = 3 {\tan}^{-} 1 \left(x + \sqrt{1 + {x}^{2}}\right)$

Substitute, x=tantheta=>theta=tan^-1x,where,thetain (- pi/2,pi/2)

$\therefore y = 3 {\tan}^{-} 1 \left(\tan \theta + \sqrt{1 + {\tan}^{2} \theta}\right)$

$= 3 {\tan}^{-} 1 \left(\tan \theta + \sec \theta\right)$

$= 3 {\tan}^{-} 1 \left(\sin \frac{\theta}{\cos} \theta + \frac{1}{\cos} \theta\right)$

$= 3 {\tan}^{-} 1 \left(\frac{1 + \sin \theta}{\cos} \theta\right) . . \to A p p l y \left(1\right)$

$= 3 {\tan}^{-} 1 \left(\frac{1 + \cos \left(\frac{\pi}{2} - \theta\right)}{\sin} \left(\frac{\pi}{2} - \theta\right)\right) \ldots \to A p p l y \left(2\right)$
=3tan^-1((2cos^2(pi/4-theta/2))/(2sin(pi/4-theta/2)cos(pi/4- theta/2)))

$= 3 {\tan}^{-} 1 \left(\cos \frac{\frac{\pi}{4} - \frac{\theta}{2}}{\sin} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)\right)$

$= 3 {\tan}^{-} 1 \left(\cot \left(\frac{\pi}{4} - \frac{\theta}{2}\right)\right)$

$= 3 {\tan}^{-} 1 \left(\cot \left\{\frac{\pi}{2} - \frac{\pi}{4} - \frac{\theta}{2}\right\}\right)$

=3tan^-1(cot(pi/2-(pi/4+theta/2))...toApply(3)

$= 3 {\tan}^{-} 1 \left(\tan \left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right) . . \to A p p l y \left(4\right)$

Now,

$\theta \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$\implies \frac{\theta}{2} \in \left(- \frac{\pi}{4} , \frac{\pi}{4}\right) \ldots \to$[Dividing by $2$]

$\implies \frac{\pi}{4} + \frac{\theta}{2} \in \left(- \frac{\pi}{4} + \frac{\pi}{4} , \frac{\pi}{4} + \frac{\pi}{4}\right) \ldots \to$[Adding $\frac{\pi}{4}$]

$\implies \frac{\pi}{4} + \frac{\theta}{2} \in \left(0 , \frac{\pi}{2}\right) \subset \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

So,

$y = 3 \left(\frac{\pi}{4} + \frac{\theta}{2}\right)$

$y = \frac{3 \pi}{4} + \frac{3}{2} \cdot \theta , w h e r e , \theta = {\tan}^{-} 1 x$

$\implies y = \frac{3 \pi}{4} + \frac{3}{2} {\tan}^{-} 1 x$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 0 + \frac{3}{2} \left(\frac{1}{1 + {x}^{2}}\right)$

i.e. (dy)/(dx)=3/(2(1+x^2)