# let f(x)= `sqrt(sin(x)+cos(x))` , determine to an accuracy of 4 decimal places: `(d^3f(2))/dx^3`

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The function f(x) = `sqrt(sin x+cos x)`

f(x) = `sqrt(sin x +cos x)`

f'(x) = `-(sin x -cos x )/(2*sqrt(sin x+cos x)) `

Use the quotient rule:`f(x) = (g(x))/(h(x)) => ``f'(x) = (g'(x)*h(x) - g(x)*h'(x))/(h(x))^2`

f''(x) = `(-sqrt(sin x+cos x)*(3+2*cos x *sin x))/(4 + 8*cosx*sinx) `

f'''(x) = `-(5*sin^3 x - 7*cos x*sin^2x + 7*cos^2x*sin x - 5*cos^3 x )/(sqrt(sin x+cos x)*(8+ 16*cos x*sin x)) `

f'''(2) = -5.58475, [Note: The value of x = 2 radians]

**The value of `f'''(2) ~~ -5.58475` **

let f(x)= sqrt(sin(x)+cos(x))

Given

`f(x)=sqrt(sin(x)+cos(x))`

`cos(x)=sin(pi/2+x)`

`sin(x)+cos(x)=sin(x)+sin(pi/2+x)`

`=2sin(pi/4+x/2)cos(pi/4)`

`=sqrt(2)sin(pi/4+x/2)`

`f(x)=(sqrt(2)sin(pi/4+x/2))^(1/2)`

`f(x)=(2)^(1/4)sin(pi/4+x/2)^(1/2)`

`f'(x)=(1/2)(2)^(1/4)cos(pi/4+x/2)/sqrt(sin(pi/4+x/2))`

`f''(x)=-(2)^(-5/4)(sin^2(pi/4+x/2)+1)/(sin(pi/4+x/2))^(3/2)`

`f'''(x)=(2)^(-7/4){cos(pi/4+x/2)(sin^2(pi/4+x/2)+3)}/(sin^4(pi/4+x/2))`

`f'''(2)=.5xx.5946{-.212958(3.954649)}(1.097268)`

`=-.27473`

`Ans.`

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