# Differentiation Rules

• Nov 18th 2012, 11:21 PM
fantasylo
Differentiation Rules
Hi, I do not understand the rule 5. Is it possible to include an example for me to understand better?

Thanks.
Attachment 25804
• Nov 18th 2012, 11:30 PM
Prove It
Re: Differentiation Rules
That's the chain rule. It's easier to write it like this: \displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align*}. Basically it means that if you have a composition of functions, you need to isolate the "inner" function (u) and work out its derivative, then write y as a function of u and evaluate its derivative. After converting it back to a function of x, multiply the two derivatives together to get the total derivative.
• Nov 18th 2012, 11:44 PM
MarkFL
Re: Differentiation Rules
Here are a couple of examples:

a) $\displaystyle y=\sin(x^3)$

$\displaystyle \frac{dy}{dx}=\cos(x^3)\frac{d}{dx}(x^3)=3x^2\cos( x^3)$

b) $\displaystyle y=e^{\tan(2x)}$

$\displaystyle \frac{dy}{dx}=e^{\tan(2x)}\frac{d}{dx}(\tan(2x))=e ^{\tan(2x)}\sec^2(2x)\frac{d}{dx}(2x)=2\sec^2(2x)e ^{\tan(2x)}$
• Nov 19th 2012, 08:00 AM
Prove It
Re: Differentiation Rules
Quote:

Originally Posted by MarkFL2
Here are a couple of examples:

a) $\displaystyle y=\sin(x^3)$

$\displaystyle \frac{dy}{dx}=\cos(x^3)\frac{d}{dx}(x^3)=3x^2\cos( x^3)$

b) $\displaystyle y=e^{\tan(2x)}$

$\displaystyle \frac{dy}{dx}=e^{\tan(2x)}\frac{d}{dx}(\tan(2x))=e ^{\tan(2x)}\sec^2(2x)\frac{d}{dx}(2x)=2\sec^2(2x)e ^{\tan(2x)}$

Or to use the easier-to-remember form of the chain rule that I posted...

1. \displaystyle \displaystyle \begin{align*} y = \sin{\left( x^3 \right)} \end{align*}

Let \displaystyle \displaystyle \begin{align*} u = x^3 \end{align*} which gives \displaystyle \displaystyle \begin{align*} y = \sin{u} \end{align*}.

Then \displaystyle \displaystyle \begin{align*} \frac{du}{dx} = 3x^2 \end{align*} and \displaystyle \displaystyle \begin{align*} \frac{dy}{du} = \cos{u} = \cos{\left(x^3 \right)} \end{align*}.

So \displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3x^2\cos{\left(x^3 \right)} \end{align*}

See if you can use a similar process with the second example. You will need to use the Chain Rule TWICE.
• Nov 19th 2012, 09:46 PM
fantasylo
Re: Differentiation Rules
Okay. I got it. Thank you Prove It and MarkFL2 for your clear examples and explanation!!