F(x) = 7x^2 + 4, find and simplify using the derivative function f(a+h) -f(a)/ (h) ?

2 Answers
Aug 23, 2017

#(dF)/(dx) = 14x#

Explanation:

I'm not 100% sure what you are asking but I'm going to take a punt that you want to find the derivative of #F(x)# using the limit definition? You have left the limit out of your 'derivative function' but this is what I think you meant.

#(df)/(da) = lim_(h->0) (f(a+h) - f(a))/h#

With #F(x) = 7x^2 + 4# we have

#(dF)/(dx) = lim_(h->0) (7(x+h)^2 + 4 - (7x^2+4))/h#

Expanding out the squared bracket some terms helpfully cancel out:

#=lim_(h->0) (color(red)(7x^2) + 14xh + 7h^2 + color(blue)(4) color(red)(- 7x^2) color(blue)(- 4))/h#

#=lim_(hrarr0) 14x + 7h = 14x#

#therefore (dF)/(dx) = 14x#

Pleasingly this is what we would expect had we differentiated normally using the power rule.

Aug 23, 2017

We use the expression #f'(x) = lim_(h->0) (f(x + h) - f(x))/h#.

#f'(x) =lim_(h->0) (7(x + h)^2 + 4 - (7x^2 + 4))/h#

#f'(x) = lim_(h->0) (7x^2 + 14xh + h^2 + 4 - 7x^2 - 4)/h#

#f'(x) = lim_(h->0) (14xh + h^2)/h#

#f'(x) = lim_(h->0) (h(14x + h))/h#

#f'(x) = lim_(h->0) 14x + h#

#f'(x) = 14x#

This matches what we would obtain using the power rule, where #f'(x) = 7(2)x^(2 - 1) = 14x^1 = 14x#.

Hopefully this helps!