Derivative of Exponential and Logarithmic Functions

We know that exponential functions come in the follwing form:

[latex]\large{y = (a)^x}[/latex]

The graph shown illustrates a general exponential function. What would the derivative of this be? Take a look at the definition below

Derivative of Exponential Functions:

If [latex]\large{f(x) = (a)^x}[/latex]

then

[latex]f'(x) = a^x \cdot ln(a)[/latex]

Logarithmic functions, as we know, come in the form:

[latex]y = log_{a}x[/latex]

And also the natural logarithmic function:

[latex]y = ln(x)[/latex]

See the graph shown, which illustrates the general form of logarithmic functions.

Let us now use implicit differentiation to find the derivative of logarithmic functions as such.

First, if we have [latex]\large{y = \log_{a}x}[/latex] then:

[latex]a^y = x[/latex]

Now, using the method of implicit differentiation, and taking the derivative of both sides with respect to x, we have:

[latex]\large{\frac{d}{dx}(a^y) = \frac{d}{dx}(x)}[/latex]

[latex]\large{\frac{d}{dx}(a^y)\frac{dy}{dx} = 1}[/latex]

[latex]\large{(a^y \cdot ln(a))(\frac{dy}{dx}) = 1}[/latex]

[latex]\large{\frac{dy}{dx} = \frac{1}{a^y(ln(a))} = \frac{1}{xln(a)}}[/latex]

So, this tells us that:

[latex]\large{\frac{d}{dx}(log_{a}x) = \frac{1}{xln(a)}}[/latex]

Then, to get the derivative of the natural log function, we will just substitute e = a in the above derivative, as shown:

[latex]\large{\frac{d}{dx}(log_{e}x) = \frac{d}{dx}(ln(x)) = \frac{1}{xln(e)} = \frac{1}{x}}[/latex]

This shows that:

[latex]\large{\frac{d}{dx}(ln(x)) = \frac{1}{x}}[/latex]

[latex]\large{\frac{d}{dx}(log_{a}x) = \frac{1}{xlna}}[/latex]

and

[latex]\large{\frac{d}{dx}ln(x) = \frac{1}{x}}[/latex]

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