# Differentiating and Integrating

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.

Note by Jack Rawlin
5 years, 8 months ago

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