what is the formula for calculating the instantaneous rate of change for f(x)=e^x where x=2

Question

I need to find the instantaneous rate of change for f(x) = e^x, x=2

Answer 1

Given exponential functon f(x) = e^x

Average rate of change = [f(x+h)-f(x)]/h

= [e^(x+h)-e^x]/h

= [e^xe^h-e^x]/h

= [e^x(e^h-1)]/h

The instantaneous rate of change = lim(h--->0)e^h-1/h = 1

= [e^(2+h)-e^2]/h

Remeber theformula a^(m+n) = a^m+a^n

= [e^2e^h-e^2]/h

= e^2(e^h-1)/h

instantaneous rate of change for f(x) =e^x, x = 2

Lim(h-->0)e^h-1/h = 1

= e^2(1)

= 7.38905