Finding the Derivative.

[Johnwill]

hello,

i need some assistance on this problem:

y = e^(1+lnx)

1st. I brought down the (1+lnx) by using natural log on both sides.

lny*y'=(1+lnx)*lne*1/x
y'/y=(1/x)*1*-1/x^2

y'=e^(1+lnx)*-1/x^3

what do i do next?

 

[pka]

\(\displaystyle \L y = e^{\left( {1 + \ln (x)} \right)} = \left( e^1 \right)\left( {e^{\ln (x)} } \right) = \left( e \right)\left( x \right)\)

 

[Johnwill]

how did you get that e^lnx = x

 

[pka]

how did you get that e^lnx = x

My goodness, you don't understand the relation of e<SUP>x</SUP> and ln(x).
You need to read you textbook!

 

[Johnwill]

nvm, gotcha
Thanks for the warning?