So we all have a basic idea of differentiating rules, for example; both of these hold true

\[y = Log_{e}{x} \Rightarrow \frac{dy}{dx} = \frac{1}{x}\]

\[y = e^{x} \Rightarrow \frac{dy}{dx} = e^{x}\]

But why?

Well it doesn't take much to show both are true. Let's start with \(y = Log_{e}{x}\)

\[y = Log_{e}{x}\]

First let's use \(e\) to the power of both sides to get rid of that \(Log_{e}\)

\[e^{y} = x\]

Then use the chain rule and implicit differentiating to get

\[\frac{dy}{dx} \cdot e^{y} = 1\]

Rearrange to make \(\frac{dy}{dx}\) the subject

\[\frac{dy}{dx} = \frac{1}{e^{y}}\]

Substitute \(Log_{e}{x}\) for \(y\)

\[\frac{dy}{dx} = \frac{1}{e^{Log_{e}{x}}}\]

Cancel the \(e\) and \(Log_{e}\) to get

\[\frac{dy}{dx} = \frac{1}{x}\]

So that's that, but what about \(y = e^{x}\)?

\[y = e^{x}\]

First we'll use a natural \(Log\) on both sides

\[Log_{e}{y} = x\]

Then differentiate implicitly

\[\frac{dy}{dx} \cdot \frac{1}{y} = 1\]

Make \(\frac{dy}{dx}\) the subject

\[\frac{dy}{dx} = y\]

Substitute \(e^{x}\) for \(y\)

\[\frac{dy}{dx} = e^{x}\]

That's those two taken care of.

What if we integrate \(Log_{e}{x}\)?

\[\int_{}{} Log_{e}{x}\]

First we need to find a way to \(not\) integrate \(Log_{e}{x}\). I know let's use integration by parts!

\[\int_{}{} Log_{e}{x} \cdot 1\]

Good, let's make \(u = Log_{e}{x}\) and \(\frac{dv}{dx} = 1\)

We know:

\[\int_{}{} u\frac{dv}{dx} = uv - \int_{}{} v\frac{du}{dx}\]

So let's work it out then.

\[v = x\]

\[\frac{du}{dx} = \frac{1}{x}\]

\[\int_{}{} Log_{e}{x} = xLog_{e}{x} - \int_{}{} 1\]

\[\int_{}{} Log_{e}{x} = xLog_{e}{x} - x + c\]

\[\int_{}{} Log_{e}{x} = x(Log_{e}{x} - 1) + c\]

That's all for now.

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