derivative of `sqrt(cosx)` by first principle

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aruv | High School Teacher | (Level 2) Valedictorian

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We have to find derivative of the function `f(x)=sqrt(cos(x))` by first principle.

`f(x)=sqrt(cos(x))`                 (i)

`f(x+Deltax)=sqrt(cos(x+Deltax))`      (ii)

Subtract (i) from (ii), we get

`f(x+Deltax)-f(x)=sqrt(cos(x+Deltax))-sqrt(cos(x))`

`=(sqrt(cos(x+Deltax))-sqrt(cos(x)))(sqrt(cos(x+Deltax))+sqrt(cos(x)))/(sqrt(cos(x+Deltax))+sqrt(cos(x)))`

`=(cos(x+Deltax)-cos(x))/(sqrt(cos(x+Deltax))+sqrt(cos(x)))`

`=(2sin(x+(Deltax)/2)sin(-(Deltax)/2))/(sqrt(cos(x+Deltax))+sqrt(cos(x)))`

`lim_(Deltax->0)(f(x+Deltax)-f(x))/2=f'(x)=lim_(Deltax->0)(2sin(x+(Deltax)/2)sin(-(Deltax)/2))/(Deltax(sqrt(cos(x+Deltax))+sqrt(cos(x))))`

`=lim_(Deltax->0)(sin(x+(Deltax)/2)/(sqrt(cos(x+Deltax))+sqrt(cos(x)))xx(-sin((Deltax)/2))/((Deltax)/2))`

`=sin(x)/(sqrt(cos(x))+sqrt(cos(x)))xx(-1)`

`=(-sin(x))/(2sqrt(cos(x)))`

`Thus`

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

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