How do you find #(f(x+Δx)-f(x))/(Δx)# if #f(x) = 8x^2+1#?

1 Answer
Apr 24, 2018

# (f(x+Delta x)-f(x))/(Delta x) = 16x + 8Delta x #

Leading to the derivative:

# f'(x) = 16 x#

Explanation:

We seek a simplification of the expression:

# F(x) = (f(x+Delta x)-f(x))/(Delta x)#, say, where #f(x) = 8x^2+1#

So we evaluate:

# F(x) = ((8(x+Delta x)^2+1)-(8x^2+1))/(Delta x) #

# \ \ \ \ \ \ \ = (8(x^2+2xDelta x + (Delta x^2)+1)-(8x^2+1))/(Delta x) #

# \ \ \ \ \ \ \ = (8x^2+16xDelta x + 8(Delta x)^2+1-8x^2-1)/(Delta x) #

# \ \ \ \ \ \ \ = (16xDelta x + 8(Delta x^2))/(Delta x) #

# \ \ \ \ \ \ \ = 16x + 8Delta x #

The astute reader will not that the given expression is the Newton difference quotient , as such we can readly compute (from first principles) the derivative, #f'(x)# of the given function #f(x)#, thus:

# f'(x) = lim_(Delta x rarr 0) F(x) #

# \ \ \ \ \ \ \ \ = lim_(Delta x rarr 0) 16x + 8Delta x #

# \ \ \ \ \ \ \ \ = 16x + 0 #

# \ \ \ \ \ \ \ \ = 16x #