# 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
2 years, 5 months ago

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