derivative of a function

Question

find the derivative of f(x) = √x+1 at x = 0

 

the derivative is 1/2x², but i want to know the long way f(x+h)/h-f(x)/h method, thanks in advance

Answer 1

Hi Ragmar,

 

To be clear, the derivative of x^½ + 1  is ½x^-½, it isn't ½x²

 

Now then, to have some fun.  I'm going to let u(s) stand in for s^(1/2), because it's easy to type. Note that

[u(s)]² = s

at least on the non-negative reals, where the sqrt function is defined.

 

The derivative is given by

 

d(x^½ + 1)   =   lim        u(x+h) + 1 - ( u(x) + 1)

     dx               h→0                h

               

                  =   lim      u(x+h) - u(x)

                       h→0           h

 

I'm going to multiply top and bottom of this fraction by u(x+h) + u(x)

 

                  = lim           [u(x+h) - u(x)] [u(x+h) + u(x)]
                     h→0                      h         [u(x+h) + u(x)]

 

                  = lim           [u(x+h)]² - [u(x)]²
                    h→0          h [u(x+h) + u(x)]

 

Recalling our rule that [u(s)]²

                  = lim           [x+h] - x      
                    h→0    h [u(x+h) + u(x)]

 

                  = lim                 h                 
                    h→0     h [u(x+h) + u(x)]

 

                  = lim                1                
                     h→0      [u(x+h) + u(x)]

 

Now, as h→0, u(x+h)→u(x)

 

                 =         1          
                       u(x) + u(x)

 

                 = __1__
                    2u(x)

 

                 = ½ 1/(√x)

 

I hope that helps.

 

Regards,

Mike N.

Answer 2

f(x) =√x+1

 

f' = lim as h ->0 f(x+h)/h - f(x)/h

 

=(√x+h+1 - √x+1)/h

 

now multipy numerator and denominator of the expression above by (√x+h+1 + √x+1)

 

= x+h+1 -x-1/h(√x+h+1 + √x+1)

 

= h/h(√x+h+1 + √x+1)

 

= 1/(√x+h+1 + √x+1)

 

as h goes to zero this become

 

= 1/2√x+1      and this is the deravitive

 

note this is the correct derivative and not 1/2x² you mentioned in your question