Question

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

Answer 1

`(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=3tan^-1(x+sqrt(1+x^2))`

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

`:.y=3tan^-1(tantheta+sqrt(1+tan^2theta))`

`=3tan^-1(tantheta+sectheta)`

`=3tan^-1(sintheta/costheta+1/costheta)`

`=3tan^-1((1+sintheta)/costheta)..to Apply(1)`

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

`=3tan^-1(cos(pi/4-theta/2)/sin(pi/4-theta/2))`

`=3tan^-1(cot(pi/4-theta/2))`

`=3tan^-1(cot{pi/2-pi/4-theta/2})`

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

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

Now,

`theta in (-pi/2,pi/2)`

`=>theta/2 in(-pi/4,pi/4)...to`[Dividing by `2`]

`=>pi/4+theta/2 in (-pi/4+pi/4,pi/4+pi/4)...to`[Adding `pi/4`]

`=>pi/4+theta/2 in(0,pi/2)sub(-pi/2, pi/2)`

So,

`y=3(pi/4+theta/2)`

`y=(3pi)/4+3/2*theta,where, theta=tan^-1x`

`=>y=(3pi)/4+3/2tan^-1x`

`=>(dy)/(dx)=0+3/2(1/(1+x^2))`

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