# Proof of Derivative of Exponential function by eliminating the Exponential notation

The derivative rule of exponential function $a$ raised to the power of $x$ is derived fundamentally from the first principle of differentiation.

$\dfrac{d}{dx}{\big(a^x}\big)$ $\,=\,$ $a^x} \times \log_{e}{a$

It can also be proved in another method. So, let’s learn how to prove the differentiation of exponential function by eliminating the exponential form.

### Covert the Exponential equation into Logarithm

Proving the differentiation of an exponential function directly is not possible in differential calculus but removing the exponential notation is the only way for overcoming it.

Assume that $y$ $\,=\,$ $a^x$

Logarithms is the recommendable technique to eliminate the equation from the exponential form. So, take natural logarithms on both sides of the equation.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $\log_{e}{\big(a^x}\big)$

$\implies$ $\log_{e}{y}$ $\,=\,$ $\log_{e}{\big(a^x}\big)$

There is nothing to simplify in the expression on left-hand side of the equation. Now, focus on the expression on the right-hand side of the equation. The log of an exponential function can be simplified by the power rule of logarithms.

$\implies$ $\log_{e}{y}$ $\,=\,$ $x \times \log_{e}{(a)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{e}{y}$ $\,=\,$ $x \times \log_{e}{a}$

### Find the Differentiation of Logarithmic equation

The logarithmic equation is defined in terms of a variable $x$. Now, differentiate each side expression with respect to $x$ to perform the differentiation.

$\implies$ $\dfrac{d}{dx}{\big(\log_{e}{y}\big)}$ $\,=\,$ $\dfrac{d}{dx}{\big(x \times \log_{e}{a}\big)}$

On left hand side of the equation, the natural logarithm of a variable $y$ should be calculated with respect to another variable $x$. The variable $y$ cannot be considered as a constant. So, the natural logarithm of a variable $y$ is a function and its differentiation can be performed by the chain rule.

The differentiation of natural logarithmic function is calculated by the derivative rule of logarithms while using the chain rule.

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{\big(y\big)}$ $\,=\,$ $\dfrac{d}{dx}{\big(x \times \log_{e}{a}\big)}$

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\dfrac{d}{dx}{\big(x \times \log_{e}{a}\big)}$

Now, it is time to concentrate on finding the differentiation on the right hand side of the equation. The derivative of product of two functions can be calculated by using the product rule of derivatives.

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $x \times \dfrac{d}{dx}{\big(\log_{e}{a}\big)}$ $+$ $\log_{e}{a} \times \dfrac{d}{dx}{\big(x\big)}$

The quantity $a$ and base $e$ are constants in the logarithmic factor. So, its value is a constant. According to the differentiation rule of a constant, the derivative of a constant with respect to variable $x$ is zero. As per the differentiation rule of a variable, the derivative of $x$ with respect to $x$ is one.

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $x \times 0$ $+$ $\log_{e}{a} \times 1$

The differentiation is successfully completed and it is time to simplify the whole equation.

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $0$ $+$ $\log_{e}{a}$

$\implies$ $\dfrac{1}{y} \times \dfrac{dy}{dx}$ $\,=\,$ $\log_{e}{a}$

$\implies$ $\dfrac{1 \times dy}{y \times dx}$ $\,=\,$ $\log_{e}{a}$

$\implies$ $\dfrac{dy}{y \times dx}$ $\,=\,$ $\log_{e}{a}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{dy}{dx}$ $\,=\,$ $y \times \log_{e}{a}$

### Evaluate the Derivative of Exponential function

It is evaluated that the derivative of variable $y$ with respect to $x$ is equal to the variable $y$ times natural logarithm of $a$.

$\implies$ $\dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times \log_{e}{a}$

It is time to replace the value of variable $y$. We have taken that the variable $y$ is equal to the $a$ to the $x$-th power. So, let’s replace its value on both sides of the equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\big(a^x}\big)$ $\,=\,$ $a^x} \times \log_{e}{a$

Therefore, it is derived that the derivative of the $x$-th power of $a$ with respect to $x$ is equal to the product of $a$ to the $x$-th and natural logarithm of $a$. According to the logarithms, the logarithmic factor is also written as follows.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\big(a^x}\big)$ $\,=\,$ $a^x}\ln{a$

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