how to differentiate surd X using first principle?

Question

it would be nicer if you can teach me step by step or maybe a link to a website.

Answer 1

Well, that was a nice proof using implicit differentiation, but let us see if I can offer you a more direct proof:

d(√x)/dx = [h→0]lim (√(x+h)-√x)/h

Multiply numerator and denominator by √(x+h)+√x:

[h→0]lim (x+h-x)/(h(√(x+h)+√x))

Simplify:

[h→0]lim h/(h(√(x+h)+√x))

Cancel:

[h→0]lim 1/(√(x+h)+√x)

Evaluate:

1/(√(x+0)+√x)

Simplify:

1/(2√x)

Answer 2

f(x) = 3x^2 -5 then f(x+h) = { 3(x+h)^2 - 5 [ d/dx 5 =0] d/dx 3x^2 --d/dx 5 = d/dx 3x^2 -0 now d/dx 3x^2 =limh-- >0 3{ ( x+h)^2 -x^2}/h x isn't reminiscent of 0 so limh---->0 3x^2[ {a million+h/x)^2 --a million]/h x isn't reminiscent of 0 additionally h has an inclination to 0. so we might think of hto be numerically smller than x i. e. h/x to be smaller than cohesion .we are able to as a result advance (a million+h/x)^2 via biomial theorem and write as limh---->0 3 x^2 [a million+2h/x +(h/x)^2 -a million]/h so limh-->0 3x^2 [2/x +h/x^2]= lim h--->0 3x^2 [ 2/x + (0)] d/dx [3x^2 -5] =3x^2 .2/x =6x ans

Answer 3

if you mean derivative of sqrt(x) by first principle, try the following link

Answer 4

suppose y^2 = x
(y + dy)^2 = x + dx
y^2 + 2ydy + o(dy) = x + dx
so lim ( (y+dy)^2 - y^2)/dx) = lim( 2ydy/dx) = 2y y' = 1
so y' = 1/(2y) = 1/ (2sqrt(x))