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Differentiate y=a^xusing first principle?

1 Answer

Feb 19, 2018

$\frac{d y}{d x} = a^{x} ln (a)$ ${dy}/{dx} = a^x ln(a)$

Explanation:

First, you have to know the limit

$lim_{h->0} (a^h-1)/h = ln(a)$

Now, for the problem.

$y(x) = a^x$

$y'(x) = lim_{h->0} (y(x+h)-y(x))/h$

$= lim_{h->0} (a^(x+h) - a^x)/h$

$= lim_{h->0} (a^x*a^h - a^x)/h$

$= a^x lim_{h->0} (a^h - 1)/h$

$= a^x ln(a)$

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