How to derivate these?

**Skolover** · November 16th 2010, 10:35 AM

Hello, first year at university and a couple of odd looking functions are giving me a headache

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

Anyone bother to take a look at them please?

How to derivate these?-math_image.aspx.gif

and the other one:

Thanks in advance

**TheCoffeeMachine** · November 16th 2010, 11:13 AM

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!

**Skolover** · November 16th 2010, 11:46 AM

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

**TheCoffeeMachine** · November 16th 2010, 12:04 PM

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 $(\ln(0.1))^2$ .

managed to get the other one right by following your guidelines thanks

No problem. I'm glad that I was of help.

My 3

th post.

**Skolover** · November 16th 2010, 12:10 PM

ah i see

so $\ln^2{x} = (\ln{(x}))^2$
are the same ^^ that'll come in handy :P new way to type it

Congrats on your 300th post!

**TheCoffeeMachine** · November 16th 2010, 01:51 PM

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:

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

Originally Posted by Skolover

Congrats on your 300th post!

Thanks.

**skeeter** · November 16th 2010, 01:56 PM

$e^{2\ln{x}} = e^{\ln{x^2}} = x^2$

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

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