Differentiate `y=sqrt x` using the first principles
3 Answers
justaguide | College Teacher | (Level 2) Distinguished 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)`
jeew-m | College Teacher | (Level 1) Educator Emeritus
`y = sqrt(x)` ----(1)
If x changes by a small value `deltax ` let say y changes by `deltay` .
`y+deltay = sqrt(x+deltax)` ----(2)
(2)-(1)
`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)`
Sameera W | (Level 1) eNoter
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/(√(x+0)+√x)
dy/dx=1/(2√x) this is the answer