1. ## Derivative

y= e^x / 4+ln2x

find the derivative at x=1,
f'(1)

2. Originally Posted by john-1
y=e^x/ (1+lnx)
y' at x= 1 = 0

Can someone confirm this for me?
Standard quotient rule

$\displaystyle u = e^x \: \rightarrow \: u' = e^x$

$\displaystyle v = 1+ \ln(x) \:\rightarrow\: v' = \frac{1}{x}$

$\displaystyle y' = \frac{u'v-v'u}{v^2} = \frac{e^x(1+\ln x) - \frac{e^x}{x}}{(1+\ln x)^2}$

No idea what your condition means, I'm guessing y'(1) = 0

Hence $\displaystyle y'(1) = \frac{e - e}{1} = 0$

Note that $\displaystyle y' = m$. To find a co-ordinate sub in x=1 into y (not y')

$\displaystyle f(1) = e$ so we have the co-ordinate (1,e)

You can now use the equation of a straight line which is $\displaystyle y-y_1 = m(x-x_1)$

3. find the derivative at x=1,
f'(1)

I ended up getting 0.

4. Originally Posted by john-1
find the derivative at x=1,
f'(1)

I ended up getting 0.
Which is wrong, did you use e^(i*pi)'s expression for the derivative?

If not what did you use.

CB