Please help me understand Leibniz' notation

**Paze** · February 17th 2013, 08:39 PM

For some inexplicable reason, I cannot seem to understand Leibniz' notation.

I get it at its most simple form such as: $\frac{dy}{dx}x^2$ I do see that we are taking the difference (or delta) of y which is $\left( (x+\Delta x)(x+\Delta x)-x^2\left)$ and then I take the difference (delta) of x which is $x+\Delta x-x$ and then of course, take the limit as $\Delta x \rightarrow 0$

However when we start playing around with Leibniz' notation, I am at a loss. For instance, now I am trying to understand this

Let $w=\ln$

$\frac{d}{dx}e^w=\frac{d}{dx}x=1\\\\\Rightarrow \left(\frac{d}{dw}e^w\right)\left(\frac{dw}{dx} \right)=1\\\\\Rightarrow e^w\cdot \frac{dw}{dx}=1\Rightarrow\frac{dw}{dx}=\frac{1}{e ^w}=\frac{1}{x}$

(if it helps, here are my notation for the lecture: http://i.imgur.com/1fjBycY.jpg. The problem is on the right side.)

But I am having severe issues understanding what is going on!

I do understand the calculations if presented to me without the Leibniz' notation, I am sure. But this notation seems to clog me up every time! Please help MHF!

**chiro** · February 17th 2013, 10:21 PM

Hey Paze.

Basically this is just the chain rule where for w = f(x) and g(w) = e^w and you are differentiating d/dx g(f(x)).

**Paze** · February 18th 2013, 06:15 AM

Originally Posted by chiro

Hey Paze.

Basically this is just the chain rule where for w = f(x) and g(w) = e^w and you are differentiating d/dx g(f(x)).

Thank you, I did suspect that but my problem is understanding what is going on with the notation. Why does $\frac{d}{dx}$ become $\frac{d}{dw}$ ?

**hollywood** · February 18th 2013, 10:23 AM

Don't let Leibniz's notation scare you. Whether you write $\frac{d}{dx}f(x)$ or f'(x), it's just the derivative of f.

Your professor introduced a new variable (or a new function, depending on how you look at it) $w=\ln{x}$ . So $e^w=x$ , and you take the derivative of both sides. The right side is easy. On the left, since w is a function of x, you have a composite function ( $e^w$ is $e^{\ln{x}}$ ), so you need to use the chain rule. The derivative of the outside function is $\frac{d}{dw}e^w=e^w$ , and the derivative of the inside function is $\frac{dw}{dx}$ . So you have $e^w\frac{dw}{dx}=1$ , so $\frac{dw}{dx}=\frac{1}{e^w}=\frac{1}{x}$ , and since $w=\ln{x}$ , $\frac{d}{dx}\ln{x}=\frac{1}{x}$ , which is what you wanted to show.

If you can't figure out the derivation, you should at least memorize the result $\frac{d}{dx}\ln{x}=\frac{1}{x}$ . Or in the other notation, if $f(x)=\ln{x}$ , $f'(x)=\frac{1}{x}$ .

Hope that helps.

- Hollywood

**Paze** · February 18th 2013, 05:01 PM

Originally Posted by hollywood

Don't let Leibniz's notation scare you. Whether you write $\frac{d}{dx}f(x)$ or f'(x), it's just the derivative of f.

Your professor introduced a new variable (or a new function, depending on how you look at it) $w=\ln{x}$ . So $e^w=x$ , and you take the derivative of both sides. The right side is easy. On the left, since w is a function of x, you have a composite function ( $e^w$ is $e^{\ln{x}}$ ), so you need to use the chain rule. The derivative of the outside function is $\frac{d}{dw}e^w=e^w$ , and the derivative of the inside function is $\frac{dw}{dx}$ . So you have $e^w\frac{dw}{dx}=1$ , so $\frac{dw}{dx}=\frac{1}{e^w}=\frac{1}{x}$ , and since $w=\ln{x}$ , $\frac{d}{dx}\ln{x}=\frac{1}{x}$ , which is what you wanted to show.

If you can't figure out the derivation, you should at least memorize the result $\frac{d}{dx}\ln{x}=\frac{1}{x}$ . Or in the other notation, if $f(x)=\ln{x}$ , $f'(x)=\frac{1}{x}$ .

Hope that helps.

- Hollywood

I get it!

God I love when the solution just appears to you after scouring for minutes, hours or even days. What you wrote was Hebrew to me for the first 30 minutes or so but after that it just suddenly clicked. That feeling is just #1. Thank you very much.

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