Question

Find the derivative of the following function f(x) w.r.t. x, at x = 1 : 

`f(x)=cos^-1[sin sqrt((1+x)/2)]+x^x`

Answer 1

`f(x)=cos^-1[sin sqrt((1+x)/2)]+x^x`

            f₁(x)                            f₂(x)

`Let  f_1(x)=cos^−1 [sin sqrt((1+x)/2)] and f_2(x)=x^x`

Now,

`f_1(x)=cos^−1 [sin sqrt((1+x)/2)]`

`= f_1(x)=cos^−1 [cos(pi/2- sqrt((1+x)/2))]`

`=pi/2- sqrt((1+x)/2)`

`⇒f_1'(x)=−1/2sqrt(2/(1+x))=−sqrt(1/(2(1+x)))`

and

`f_2(x)=x^x` Taking log on both sides, we get

`log f_2(x)=xlogx`

`⇒1/(f_2(x)) f2′(x)=logx+x⋅1/x`

`=>1/(f_2(x)) f2′(x)=logx+1`

`=>f2′(x)=f_2(x)(logx+1)`

`=>f2′(x)=x^x(logx+1)`

`∵f(x)=f_1(x)+f_2(x)`

`∵f'(x)=f_1'(x)+f_2'(x)`

`=-sqrt(1/(2(1+x)))+x^x (logx+1)`

At x=1

`f'(1)=-sqrt(1/(2(1+1)))1^1(log1+1)`

`=-1/2+1=1/2`