Derivatives Of Logarithms And Exponentials

In this section we will understand more about the derivatives of exponentials and logarithms.

I. Derivative Of e^x:

By definition of derivative,

$\dfrac{d}{dx}e^x$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{e^{x + \Delta x} - e^x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{e^x \cdot e^{\Delta x} - e^x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{e^x ( e^{\Delta x} - 1)}{\Delta x}$

$= e^x \lim\limits_{\Delta x \rightarrow 0} \dfrac{e^{\Delta x} - 1}{\Delta x}$

$= e^x \cdot \ln e $ (using $\lim\limits_{x \rightarrow 0} \dfrac{a^x - 1}{x} = \ln a$ from )

$= e^x \cdot 1$

$= e^x$

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We will take a look at another derivation of this using the basic expansion of $e^x$

We know from the properties of that:

$e^x = \sum \limits_{k = 0}^{\infty} \dfrac{x^k}{k!}$

$\Rightarrow e^x = \dfrac{x^0}{0!} + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} +... \infty$

$\Rightarrow e^x = 1 + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ... \infty$

Differentiating both sides w.r.t $x$:

$\dfrac{d}{dx}(e^x) = 0 + \dfrac{1 \times x^0}{1!} + \dfrac{2 \times x^1}{2!} + \dfrac{3 \times x^2}{3!} ... \infty$

$\Rightarrow \dfrac{d}{dx}(e^x) = 1 + \dfrac{\cancel{2} \times x^1}{\cancel{2} \times 1!} + \dfrac{\cancel{3} \times x^2}{\cancel{3} \times 2!} ... \infty$

$\Rightarrow \dfrac{d}{dx}(e^x) = 1 + \dfrac{x^1}{1!} + \dfrac{x^2}{2!} ... \infty$

$\Rightarrow \dfrac{d}{dx}(e^x) = \sum \limits_{k = 0}^{\infty} \dfrac{x^k}{k!}$

The series on the RHS is the same series that we started with, which is the expansion of $e^x$.

Therefore,

$\dfrac{d}{dx}(e^x) = e^x$

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II. Derivative Of $a^x$

$\dfrac{d}{dx} a^x = \lim\limits_{\Delta x \rightarrow 0} \dfrac{a^{x + \Delta x} - a^x}{\Delta x}$

$= a^x \lim\limits_{\Delta x \rightarrow 0} \dfrac{a^{\Delta x} - 1}{\Delta x}$

$= a^x \ln a$

III. Derivative Of $\ln (x)$

$\dfrac{d}{dx} \ln x = \lim\limits_{\Delta x \rightarrow 0} \dfrac{\ln (x + \Delta x) - \ln x}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{\ln \left(\dfrac{x + \Delta x}{x}\right)}{\Delta x}$

$= \lim\limits_{\Delta x \rightarrow 0} \dfrac{1}{x} \cdot \dfrac{\ln \left(1 + \dfrac{\Delta x}{x}\right)}{\dfrac{\Delta x}{x}}$

$= \dfrac{1}{x} \cdot \lim\limits_{\frac{\Delta x}{x} \rightarrow 0} \dfrac{\ln \left(1 + \dfrac{\Delta x}{x}\right)}{\dfrac{\Delta x}{x}}$ (because as $\Delta x \rightarrow 0$, $\dfrac{\Delta x}{x} \rightarrow 0$

$= \dfrac{1}{x} \cdot \ln e = \dfrac{1}{x} \cdot 1 = \dfrac{1}{x}$ (using $\lim\limits_{x \rightarrow 0} \dfrac{\log_a(1 + x)}{x} = \ln a$ from )

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IV. Derivative Of $\log_a x$

$\dfrac{d}{dx} \log_a x$

$= \dfrac{d}{dx} \log_e x \cdot \log_a e$

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

$= \dfrac{1}{\ln a} \dfrac{d}{dx} \ln x$

$= \dfrac{1}{x \ln a}$

V. Derivative Of $x^x$

Let us take $y = x^x$, therefore,

$\ln y = x \ln x$

Differentiating both sides of the equation w.r.t $x$, we get:

$\dfrac{d}{dx} \ln y = \dfrac{d}{dx} x \ln x$

$\Rightarrow \dfrac{d}{dy} \ln y \cdot \dfrac{d}{dx} y = \ln x + x \dfrac{d}{dx} \ln x$

$\Rightarrow \dfrac{1}{y} \cdot \dfrac{d}{dx} y = \ln x + x \cdot \dfrac{1}{x}$

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$\Rightarrow \dfrac{d}{dx}y = y(\ln x + 1)$

$\Rightarrow \dfrac{d}{dx}x^x = x^x(\ln x + 1)$