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=exdydx=exy = 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=Logexy = Log_{e}{x}

y=Logexy = Log_{e}{x}

First let's use ee to the power of both sides to get rid of that LogeLog_{e}

ey=xe^{y} = x

Then use the chain rule and implicit differentiating to get

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

Rearrange to make dydx\frac{dy}{dx} the subject

dydx=1ey\frac{dy}{dx} = \frac{1}{e^{y}}

Substitute LogexLog_{e}{x} for yy

dydx=1eLogex\frac{dy}{dx} = \frac{1}{e^{Log_{e}{x}}}

Cancel the ee and LogeLog_{e} to get

dydx=1x\frac{dy}{dx} = \frac{1}{x}

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

y=exy = e^{x}

First we'll use a natural LogLog on both sides

Logey=xLog_{e}{y} = x

Then differentiate implicitly

dydx1y=1\frac{dy}{dx} \cdot \frac{1}{y} = 1

Make dydx\frac{dy}{dx} the subject

dydx=y\frac{dy}{dx} = y

Substitute exe^{x} for yy

dydx=ex\frac{dy}{dx} = e^{x}

That's those two taken care of.

What if we integrate LogexLog_{e}{x}?

Logex\int_{}{} Log_{e}{x}

First we need to find a way to notnot integrate LogexLog_{e}{x}. I know let's use integration by parts!

Logex1\int_{}{} Log_{e}{x} \cdot 1

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

We know:

udvdx=uvvdudx\int_{}{} u\frac{dv}{dx} = uv - \int_{}{} v\frac{du}{dx}

So let's work it out then.

v=xv = x

dudx=1x\frac{du}{dx} = \frac{1}{x}

Logex=xLogex1\int_{}{} Log_{e}{x} = xLog_{e}{x} - \int_{}{} 1

Logex=xLogexx+c\int_{}{} Log_{e}{x} = xLog_{e}{x} - x + c

Logex=x(Logex1)+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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