Find the derivative of #x + sqrtx# using first principles, the long way. Help?

2 Answers
Nov 23, 2017

The "first principles" definition of a derivative is:

#f'(x) = lim_(h to 0) (f(x+h)-f(x))/h#

We are given:

#f(x) = x + sqrtx#

To obtain #f(x+h)#, substitute #x+h# for every x in #f(x)#:

#f(x+h) = x+h + sqrt(x+h)#

Substitute #f(x+h)# and #f(x)# into the definition:

#f'(x) = lim_(h to 0) (x+h + sqrt(x+h)-(x + sqrtx))/h#

Distribute the negative sign:

#f'(x) = lim_(h to 0) (x+h + sqrt(x+h)-x - sqrtx)/h#

Combine like terms:

#f'(x) = lim_(h to 0) (h + sqrt(x+h) - sqrtx)/h#

Separate into two fractions:

#f'(x) = lim_(h to 0) h/h + (sqrt(x+h) - sqrtx)/h#

The first fraction, #h/h#, becomes 1:

#f'(x) = lim_(h to 0) 1 + (sqrt(x+h) - sqrtx)/h#

We can use the property #(a-b)(a+b) = (a^2-b^2)# to eliminate the radicals in the numerator; this means that we must multiply by 1 in the from of #(sqrt(x+h) + sqrtx)/(sqrt(x+h) + sqrtx)#:

#f'(x) = lim_(h to 0) 1 + (sqrt(x+h) - sqrtx)/h(sqrt(x+h) + sqrtx)/(sqrt(x+h) + sqrtx)#

Please observe that we have effectively multiplied by 1 and the radicals in the numerator are eliminated by squaring:

#f'(x) = lim_(h to 0) 1 + (x+h - x)/(h(sqrt(x+h) + sqrtx))#

The #x - x# in the numerator becomes 0 :

#f'(x) = lim_(h to 0) 1 + h/(h(sqrt(x+h) + sqrtx))#

The #h/h# becomes 1:

#f'(x) = lim_(h to 0) 1 + 1/(sqrt(x+h) + sqrtx)#

Now, we may let #h to 0#:

#f'(x) = 1 + 1/(sqrtx + sqrtx)#

Combine like terms:

#f'(x) = 1 + 1/(2sqrtx)#

Nov 24, 2017

#f'(x)=(2x+sqrtx)/(2x)#

Explanation:

We know:

#f'(x)=Lim_(h->0)(f(x+h)-f(x))/h#

#f'(x)=Lim_(h->0)((x+h+sqrt(x+h)-(x+sqrtx)))/h#

#f'(x)=Lim_(h->0)(x+h+sqrt(x+h)-x-sqrtx)/h#

#f'(x)=Lim_(h->0)(cancelcolor(red)x+h+sqrt(x+h)-cancelcolor(red)x-sqrtx)/h#

#f'(x)=Lim_(h->0)(h/h+(sqrt(x+h)-sqrtx)/h)#

We multiply the top and the bottom of the second fraction by the conjugate of the top:

#f'(x)=Lim_(h->0)(1+((sqrt(x+h)-sqrtx)(sqrt(x+h)+sqrtx))/(h(sqrt(x+h)+sqrtx)))#

#f'(x)=Lim_(h->0)(1+(x+h-x)/(h(sqrt(x+h)+sqrtx)))#

#f'(x)=Lim_(h->0)(1+(cancelcolor(red)x+hcancelcolor(red)(-x))/(h(sqrt(x+h)+sqrtx)))#

#f'(x)=Lim_(h->0)(1+h/(h(sqrt(x+h)+sqrtx)))#

#f'(x)=Lim_(h->0)(1+cancelcolor(red)h/(cancelcolor(red)h(sqrt(x+h)+sqrtx)))#

#f'(x)=Lim_(h->0)(1+1/(sqrt(x+h)+sqrtx))#

Now, when #h->0# we get:

#f'(x)=1+1/(sqrt(x+0)+sqrtx)=1+1/(sqrtx+sqrtx)#

#f'(x)=1+1/(2sqrtx)=1+sqrtx/(2x)# or:

#f'(x)=(2x+sqrtx)/(2x)#