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Find the Derivative of the Following Function - Mathematics

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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`

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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`

Concept: Derivatives of Inverse Trigonometric Functions
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