Differentiate `y=sqrt x` using the first principles

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jeew-m eNotes educator| Certified Educator

`y = sqrt(x)` ----(1)

If x changes by a small value `deltax ` let say y changes by `deltay` .

`y+deltay = sqrt(x+deltax)` ----(2)



`deltay = sqrt(x+deltax)-sqrt(x)`

`deltay = (sqrt(x+deltax)-sqrt(x))xx(sqrt(x+deltax)+sqrt(x))/(sqrt(x+deltax)+sqrt(x))`

`deltay = ((sqrt(x+deltax))^2-(sqrtx)^2)/(sqrt(x+deltax)+sqrtx)`

`deltay = (x+deltax-x)/(sqrt(x+deltax)+sqrtx)`

`(deltay)/(deltax) = ((x+deltax-x)/(sqrt(x+deltax)+sqrtx))/(deltax)`


The derivative is defined as;

`(dy)/dx = lim_(xrarr0)(deltay)/(deltax)`


`(dy)/dx = lim_(xrarr0)((x+deltax-x)/(sqrt(x+sqrtx)+deltax))/(deltax)`

`(dy)/dx = lim_(xrarr0)((deltax)/(sqrt(x+deltax)+sqrtx))/(deltax)`

`(dy)/dx = lim_(xrarr0)1/(sqrt(x+deltax)+sqrtx))`

`(dy)/dx = 1/(2sqrtx)`


So by first pricipals the derivative of `sqrtx` is `1/(2sqrtx)`


justaguide eNotes educator| Certified Educator

The function `y = sqrt x` . The derivative of any function f(x) using the first principles is given by the limit `lim_(h->0) (f(x+h) - f(x))/h` . For the function `f(x) = sqrt x` :

`lim_(h->0) (f(x+h) - f(x))/h`

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

=> `lim_(h->0) (sqrt(x+h) + sqrt x)(sqrt(x+h) - sqrt x)/(h*(sqrt(x+h) + sqrt x))`

=> `lim_(h->0) (x+h - x)/(h*(sqrt(x+h) + sqrt x))`

=> `lim_(h->0) h/(h*(sqrt(x+h) + sqrt x))`

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

Let h = 0

=> `1/(sqrt x + sqrt x)`

=> `1/(2*sqrt 2)`

This gives the derivative of `y = sqrt x` as `y' = 1/(2*sqrt x)`

Sameera W | Student

y=√x   (1)

y+delta (y)=√(x+delta (x)         (2)


 (2) -(1)

delta (y)=(√(x+delta (x))-√x)  . (√(x+delta (x)+√x))/(√(x+delta (x)+√x))

delta (y)=((x+delta (x))-(x))/√((x+delta (x))+√x)

delta (y)=delta (x)/(√(x+delta (x))+√x)

delta (y)/delta (x)=1/(√(x+delta (x))+√x)

when delta (x) goes to zero then delta (y)/delta (x)=dy/dx so that


dy/dx=1/(2√x) this is the answer






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