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Find the derivative of sec√x using first principle method.

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acidified | Student, Grade 11 | (Level 1) Honors

Posted May 28, 2012 at 10:45 AM via web

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Find the derivative of sec√x using first principle method.

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thilina-g | College Teacher | (Level 1) Educator

Posted May 28, 2012 at 11:42 AM (Answer #1)

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`y = sec(sqrt(x))`

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

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

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

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

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

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

`y' = lim_(h-gt0) (2xxsin((sqrt(x+h)+sqrt(x))/2))/(cos(sqrt(x+h)) xx cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

We can evaluate the first limit as,

`y' = (2xxsin((sqrt(x+0)+sqrt(x))/2))/(cos(sqrt(x+0)) xx cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

 

`y' = (2xxsin((sqrt(x)+sqrt(x))/2))/(cos(sqrt(x)) xx cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

`y' = (2xxtan(sqrt(x)))/(cos(sqrt(x))) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

`y' = 2xxtan(sqrt(x))sec(sqrt(x)) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

We will find the otehr limit separately as below.

` lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

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

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

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

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

When h-> 0, `h/(2xx(sqrt(x+h)+sqrt(x)) )-> 0`

Therefore, we can change the limit as below,

 

`=lim_((h/(2xx(sqrt(x+h)+sqrt(x))) )-gt0)sin((h)/(2xx(sqrt(x+h)+sqrt(x))))/(h/(2xx(sqrt(x+h)+sqrt(x)))) xx 1/(2xx(sqrt(x+h)+sqrt(x)))`

This gives,

`=lim_((h/(2xx(sqrt(x+h)+sqrt(x))) )-gt0)sin((h)/(2xx(sqrt(x+h)+sqrt(x))))/(h/(2xx(sqrt(x+h)+sqrt(x)))) xx 1/(2xx(sqrt(x+h)+sqrt(x))) = 1 xx 1/(2xx2sqrt(x))`

 

Therefore,

` lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h = 1/(4sqrt(x))`

 

 

Therefore,

`y' = 2xxtan(sqrt(x))sec(sqrt(x)) xx lim_(h->0)sin((sqrt(x+h) -sqrt(x))/2)/h`

`y' = 2xxtan(sqrt(x))sec(sqrt(x)) xx 1/(4sqrt(x)) `

This gives,

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

 

Which is the same answer you get via normal differentiation.

 

 

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