How could I prove this? Would this be using a theorem from real analysis?

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2 Answers
May 17, 2018

#"Use the definition of derivative :"#

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

#"Here we have"#

#f'(x_0) = lim_{h->0} (f(x_0 + h) - f(x_0))/h#
#g'(x_0) = lim_{h->0} (g(x_0 + h) - g(x_0))/h#

#"We need to prove that"#

#f'(x_0) = g'(x_0)#
#"or"#
#f'(x_0) - g'(x_0) = 0#
#"or"#
#h'(x_0) = 0#
#"with "h(x) = f(x) - g(x)#
#"or"#
#lim_{h->0} (f(x_0 + h) - g(x_0 +h) - f(x_0) + g(x_0))/h = 0#
#"or"#
#lim_{h->0} (f(x_0 + h) - g(x_0 + h))/h = 0#
#"(due to "f(x_0) = g(x_0)")"#

#"Now"#

#f(x_0 + h) <= g(x_0 + h)#
#=> lim <= 0 " if "h>0" and "lim >= 0" if "h < 0#

#"We made the assumption that f and g are differentiable"#
#"so "h(x) = f(x) - g(x)" is also differentiable,"#
#"so the left limit must be equal to the right limit, so"#

#=> lim = 0#
#=> h'(x_0) = 0#
#=> f'(x_0) = g'(x_0)#

May 18, 2018

I will provide a quicker solution than the one in https://socratic.org/s/aQZyW77G. For this we will have to rely on some familiar results from calculus.

Explanation:

Define #h(x) = f(x)-g(x)#

Since #f(x)\le g(x)#, we have #h(x) le 0#

At #x=x_0# , we have #f(x_0) = g(x_0)#, so that #h(x_0) = 0#

Thus #x=x_0# is a maximum of the differentiable function #h(x)# inside the open interval #(a,b)#. Thus

#h^'(x_0) = 0 implies#

#f^'(x_0)-g^'(x_0) implies #

#f^'(x_0) = g^'(x_0)#