find the derivative of secX^1/2 by using first principle method.

couldn't find the under root sign so used x^1/2 instead of under sq. root x.

### 1 Answer | Add Yours

We have to find the derivative of `sqrt(sec(x))` by first principles.

The definition of derivative is given by,

`f'(x) = lim_(h-gt0)(f(x+h)-f(x))/h`

Therefore,

if `y = sqrt(sec(x))` , then

`y' = lim_(h-gt0)(sqrt(sec(x+h))-sqrt(sec(x)))/h`

If you try to evaluate the limit straight away, it would be 0/0 which is indeterminate.

We can rearrange this,

`y' = lim_(h-gt0)(sqrt(sec(x+h))-sqrt(sec(x)))/h xx (sqrt(sec(x+h))+sqrt(sec(x)))/(sqrt(sec(x+h))+sqrt(sec(x)))`

This would give,

`y' = lim_(h-gt0)(sec(x+h)-sec(x))/(h (sqrt(sec(x+h))+sqrt(sec(x))))`

`y' = lim_(h-gt0)(1/cos(x+h)-1/cos(x))/(h (sqrt(sec(x+h))+sqrt(sec(x))))`

`y' = lim_(h-gt0)((cos(x)-cos(x+h))/(cos(x+h)cos(x)))/(h (sqrt(sec(x+h))+sqrt(sec(x))))`

This gives,

`y' = lim_(h-gt0)((-2sin((2x+h)/2)sin(-h/2))/(cos(x+h)cos(x)))/(h (sqrt(sec(x+h))+sqrt(sec(x))))`

This gives,

`y' = lim_(h-gt0)sin(x+h/2)/(cos(x+h)cos(x)(sqrt(sec(x+h))+sqrt(sec(x)))) xx lim_(h-gt0)(sin(h/2)/(h/2))`

h goes to 0 mean, h/2 alos goes to 0.Then,

`y' = sin(x)/(cos(x)cos(x)(sqrt(sec(x))+sqrt(sec(x)))) xx lim_(h/2-gt0)(sin(h/2)/(h/2))`

We know,

`lim_(h/2-gt0)(sin(h/2)/(h/2)) = 1`

Therefore,

`y' = sin(x)/(cos(x)cos(x)(sqrt(sec(x))+sqrt(sec(x)))) xx 1`

`y' = (sec(x)tan(x))/(2sqrt(sec(x))) `

This is the same answer we get via normal differentiation also

Therefore,

`(d(sqrt(sec(x))))/(dx) = (sec(x)tan(x))/(2sqrt(sec(x))) `

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes