How do you find f'(x) using the definition of a derivative #f(x)=6 x + 2 /sqrt{x}#?

#f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h#

# = lim_(hrarr0) ([6(x+h)+2/sqrt(x+h)] - [ 6x+2/sqrtx])/h#

This limit looks hard to evaluate, but we can split it into two limits:

# = lim_(hrarr0)(6(x+h)-6x)/h + lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h#

The first limit evaluates to #6#, so let's focus on the second limit.

#lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h#

This is a combination of problems you've probably seen before.

Like #lim_(hrarr0)(1/(x+h)-1/x)/h# and #lim_(hrarr0)(sqrt(x+h)-sqrtx)/h#.

We should try the methods that worked for those problems.

First, let's get a single rational expression in the numerator.

# lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h = lim_(hrarr0)((2sqrtx-2sqrt(x+h))/(sqrt(x+h)sqrtx))/(h/1)#

And now a single rational expression.

# = lim_(hrarr0)((2sqrtx-2sqrt(x+h))/(hsqrt(x+h)sqrtx))#

If we try to evaluate the limit, we still get the indeterminate form #0/0#.

I suppose we could try rationalizing the numerator to see if that helps.

# = lim_(hrarr0)((2sqrtx-2sqrt(x+h)))/(hsqrt(x+h)sqrtx) * ((2sqrtx+2sqrt(x+h)))/((2sqrtx+2sqrt(x+h)))#

# = lim_(hrarr0)((4x-4(x+h)))/(hsqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))#

# = lim_(hrarr0)(-4cancel(h))/(cancel(h)sqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))#

The numerator no longer goes to #0#!

That means we can evaluate the limit. (Then we'll add the #6# we got above to finish finding the derivative of #f#).

#lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h = lim_(hrarr0)(-4)/(sqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))#

# = (-4)/ (sqrt(x+0)sqrtx (2sqrtx+2sqrt(x+0))#

# = (-4)/ (sqrtxsqrtx (4sqrtx)) = (-1)/(sqrtx)^3#

Finally we finish

#f'(x) = lim_(hrarr0) ([6(x+h)+2/sqrt(x+h)] - [ 6x+2/sqrtx])/h#

# = lim_(hrarr0)(6(x+h)-6x)/h + lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h#

# = 6 - 1/(sqrtx)^3#