# How to derivate these?

• November 16th 2010, 10:35 AM
Skolover
How to derivate these?
Hello, first year at university and a couple of odd looking functions are giving me a headache (Headbang) I have to derivate them so i can the value when x: 0.1,0.2,0.3, but my derivation so far has given me all the wrong answers and not the right ones (Giggle)

Anyone bother to take a look at them please? (Clapping)

Attachment 19729

and the other one:

Attachment 19730

• November 16th 2010, 11:13 AM
TheCoffeeMachine
First, change the base $\displaystyle \frac{e^{-x^2}}{\ln_{2}{x}} = \frac{e^{-x^2}\ln(2)}{\ln{x}}$.

Since $\ln(2)$ is just a constant, you need to differentiate $\displaystyle \frac{e^{-x^2}}{\ln{x}}$.

If $\displaystyle y = \frac{e^{-x^2}}{\ln{x}}$, then:

$\ln{y} = \ln{e^{-x^2}-\ln\left(\ln{x}\right) = -x^2-\ln\left(\ln{x}\right)$.

Differentiate both sides to get:

$\displaystyle \left(\frac{1}{y}\right)\frac{dy}{dx} = -2x-\frac{1}{x\ln{x}} = \frac{-\left(2x^2\ln{x}+1\right)}{x\ln{x}}$

$\displaystyle \Rightarrow \frac{dy}{dx} = \frac{-\left(2x^2\ln{x}+1\right)}{x\ln{x}}y = \frac{-e^{-x^2}\left(2x^2\ln{x}+1\right)}{x\ln^2{x}}$

So your derivative is $\displaystyle \boxed{\frac{-\ln(2)e^{-x^2}\left(2x^2\ln{x}+1\right)}{x\ln^2{x}}}$.

Do the same for the second one!
• November 16th 2010, 11:46 AM
Skolover
Quote:

So your derivative is $\displaystyle \boxed{\frac{-\ln(2)e^{-x^2}\left(2x^2\ln{x}+1\right)}{x\ln^2{x}}}$.
Thanks for the fast reply, but i've never seen the ln^2, how is that imputted when i want to know the answer of 0.1? etc: 0.1 ln^2 0.1 , I just get a syn error on my calc ^^ any other way of typing it?

managed to get the other one right by following your guidelines thanks ;)
• November 16th 2010, 12:04 PM
TheCoffeeMachine
Quote:

Originally Posted by Skolover
Thanks for the fast reply, but i've never seen the ln^2, how is that imputted when i want to know the answer of 0.1? etc: 0.1 ln^2 0.1 , I just get a syn error on my calc ^^ any other way of typing it?

$\ln^2{x} = (\ln{x})^2 = \ln(x)\ln(x)$, of course. So either enter $\ln(0.1)\times \ln(0.1)$ or $(\ln(0.1))^2$.
Quote:

managed to get the other one right by following your guidelines thanks ;)
No problem. I'm glad that I was of help.

My 3(Cool)(Cool)th post.
• November 16th 2010, 12:10 PM
Skolover
ah i see ;)
so $\ln^2{x} = (\ln{(x}))^2$
are the same ^^ that'll come in handy :P new way to type it :D

Congrats on your 300th post! :D:D
• November 16th 2010, 01:51 PM
TheCoffeeMachine
I've just realised that 'do the same for the second one' was bit misleading, since taking
logarithms makes the function more complicated to differentiate than it already were.

We have $y = \sin^{-1}({e^{2\ln{x}}})[/Math] [Math] = \sin^{-1}(x^2})$. Using the chain rule gives:

$\displaystyle \frac{dy}{dx} = \frac{(x^2)'}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}} = \frac{2x}{\sqrt{1-x^4}}$.

Quote:

Originally Posted by Skolover
Congrats on your 300th post! :D:D

Thanks. :)
• November 16th 2010, 01:56 PM
skeeter
$e^{2\ln{x}} = e^{\ln{x^2}} = x^2$

... take the derivative of $\sin^{-1}(x^2)$