Problem with derivative y'=(lnˇ2 * lnˇ3 * lnˇ4 x)'= ?

**Tafka** · October 6th 2010, 11:19 AM

Hello,
I have a problem with derivative.Sorry,my topic is abit off :P,not use to doing this stuff at the computer.
$y = ln^2ln^3 ln^4 x$

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The book says that the right answer is $y'=\frac{24ln ln^3 ln^4 x}{x(ln ln^4 x)lnx}$

Thank you

**Opalg** · October 7th 2010, 08:23 AM

Originally Posted by Tafka

Hello,
I have a problem with derivative.Sorry,my topic is abit off :P,not use to doing this stuff at the computer.
$y = ln^2ln^3 ln^4 x$

}][[]]}
The book says that the right answer is $y'=\frac{24ln ln^3 ln^4 x}{x(ln ln^4 x)lnx}$

This question involves multiple use of the chain rule. To make it manageable, split it into smaller sections. Start by putting $u = \ln^3(\ln^4x)$ . Then $y = \ln^2u$ . By the chain rule, $\frac{dy}{dx} = 2\ln u * \frac1u * \frac{du}{dx} = {\displaystyle\frac{2\ln\bigl(\ln^3(\ln^4x)\bigr)} {\ln^3(\ln^4x)}}\frac{du}{dx}.$

Now you have to find $\frac{du}{dx}$ . Do that in the same way, putting $u = \ln^3v$ , where $v = \ln^4x$ , and using the chain rule to find $\frac{du}{dx}$ in terms of $\frac{dv}{dx}$ . Then use the chain rule once more to find $\frac{dv}{dx}$ . Finally, put everything together into a single expression and simplify it.

BTW this is a classic example of a case where using parentheses correctly makes things much easier to read: $y = \ln^2\ln^3 \ln^4 x$ looks like a jumble, but $y = \ln^2\bigl(\ln^3 (\ln^4 x)\bigr)$ is a lot clearer.

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