y= e^x / 4+ln2x
find the derivative at x=1,
f'(1)
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 $\displaystyle y-y_1 = m(x-x_1)$