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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`
Solution
`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`
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