# differentiation tan -1 and ln (e^x)

• Oct 17th 2009, 02:28 PM
calcbeg
differentiation tan -1 and ln (e^x)
Hi

I need to differentiate the function g(x) = tan ^-1 (x^(1/2))/ ln (e^x)

I am assuming I will use the quotient rule so

g'(x) = ln (e^x) X (1/(1+x^2))(1/2x^-1/2) dx - tan ^-1(x^(1/2)) e^x dx

it says not to simplify so am I right or am completely off track?

Thanks

Calculus Beginner
• Oct 17th 2009, 02:35 PM
Scott H
Are you sure the problem says $\displaystyle \ln e^x$? Because $\displaystyle \ln e^x=x$, the derivative simplifies to

$\displaystyle \frac{d}{dx}\tan^{-1}\sqrt{x}.$

Now we may use the Chain Rule together with the fact that

$\displaystyle \frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}.$
• Oct 17th 2009, 03:06 PM
chisigma
Quote:

Originally Posted by Scott H
... are you sure the problem says $\displaystyle \ln e^x$? Because $\displaystyle \ln e^x=x$, the derivative simplifies...

... may be that someone at this point horns in on and says something like that: you are wrongh!... everibody with a minimum of mathematical background knows that is...

$\displaystyle e^{\ln x} = x + 2\pi k i$

... some people are funny! ... no problem! (Itwasntme) ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 17th 2009, 03:15 PM
Scott H
Well, conventionally we use $\displaystyle z$ to denote complex numbers...