I assume you have been shown how to derive f' when f' = e^x.
f(x) = 5^x
f(x+h) = 5^(x+h) = 5^x * 5^h
f(x+h)-f(x) = 5^x(5^h-1)
dividing by h, we have
5^x (5^h-1)/h
= 5^x (e^(h ln 5)-1)/(h ln5) * ln5
Recall than lim(u->0) (e^u-1)/u = 1, so we wind up with
ln5 5^x
x^2 is ever so easy
f(x+h)-f(x) = (x+h)^2 - x^2
= x^2+2hx+h^2-x^2
= 2hx+h^2
divide that by h and you have
2x + h
and lim(h->0) is just 2x
Find the derivative of the following functions using the f prime of x = the limit as h approaches 0 of f(x+h) – f(x) all over h
a)ƒ(x) = x2 + 5^x
b)ƒ(x)= 1/2^x
1 answer