$\large \dfrac{d}{dx}{a^x} \,=\, a^x \log_{e}{a}$

It is called differentiation of $a$ raised to the power of $x$ with respect to $x$ formula and this differentiation rule can be derived in differential calculus on the basis of relation between limit and differentiation. Here the steps to prove the $\dfrac{d}{dx}{a^x}$ formula mathematically.

The derivative of a function with respect to $x$ can be expressed in limit form as per the mathematical relation between limit and differentiation. So, the differentiation of $a$ raised to the power of $x$ is expressed in limit form.

$\implies$ $\dfrac{d}{dx}{a^x} = \displaystyle \lim_{h \,\to\, 0}{\dfrac{a^{x+h}-a^x}{h}}$

An exponential function in numerator contains sum of two literals as its exponent. It can be divided as two multiplying factors by applying product rule of exponents.

$\implies$ $\dfrac{d}{dx}{a^x} = \displaystyle \lim_{h \,\to\, 0}{\dfrac{a^{x} \times a^{h}-a^x}{h}}$

$\implies$ $\dfrac{d}{dx}{a^x} = \displaystyle \lim_{h \,\to\, 0}{\dfrac{a^{x} \times (a^{h}-1)}{h}}$

$\implies$ $\dfrac{d}{dx}{a^x} = \displaystyle \lim_{h \,\to\, 0}{a^{x} \times \dfrac{a^{h}-1}{h}}$

The multiplying factor $a^x$ is a constant in this case. So, it can be taken out from the expression reasonably.

$\implies$ $\dfrac{d}{dx}{a^x} = a^{x} \times \displaystyle \lim_{h \,\to\, 0}{\dfrac{a^{h}-1}{h}}$

According to limit rules, the lim of $\dfrac{a^x-1}{x}$ as $x$ approaches $0$ is equal to natural logarithm of $a$.

$\displaystyle \lim_{h \,\to\, 0}{\dfrac{a^{h}-1}{h}} = \log_{e}{a}$

Now simplify the equation for deriving the derivative of $a^x$ with respect to $x$ formula in differential calculus.

$\implies$ $\dfrac{d}{dx}{a^x} = a^{x} \times \log_{e}{a}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{a^x} = a^{x} \log_{e}{a}$

It is used as formula to deal differentiation of $a^x$ with respect to $x$ in calculus. It can be simply written in natural logarithmic form as follows.

$\dfrac{d}{dx}{a^x} = a^{x} \ln{a}$

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