Question #e6265

1 Answer
May 2, 2017

#2x-1/x^2#

Explanation:

The definition of the derivative of a function #f(x)# is given by #lim_(h->0)(f(x+h)-f(x))/h#.

For this function, its derivative is #lim_(h->0)(((x+h)^2+1/(x+h)+7)-(x^2+1/x+7))/h=lim_(h->0)((x+h)^2+1/(x+h)+7-x^2-1/x-7)/h=lim_(h->0)((x+h)^2+1/(x+h)-x^2-1/x)/h#.

Expand #(x+h)^2# and combine #1/(x+h)-1/x# into a single function. This gives #lim_(h->0)(x^2+2xh+h^2-x^2+(x-x-h)/(x(x+h)))/h=lim_(h->0)(2xh+h^2-h/(x(x+h)))/h#.

Some of the #h#'s cancel out: #lim_(h->0)2x+h-1/(x(x+h))#.

Now, we can substitute #h=0#. This gives our final answer: #2x-1/x^2#.