How do you find the derivative of #y= e^(ex) ln x#?

Question

3103 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-05T10:39:37

#d/dx (e^(ex)lnx) = e^(ex)(elnx +1/x)#

(1) #d/dx (e^(ex)lnx) = d/dx ( e^(ex))lnx +e^(ex)d/dx(lnx)#

Now we have using the chain rule:

#d/dx ( e^(ex)) = e^(ex) d/dx (ex) = e*e^(ex)#

while:

#d/dx(lnx) = 1/x#

is a standard derivative.

Substituting this in (1) we have:

#d/dx (e^(ex)lnx) = e*e^(ex)lnx +e^(ex)/x#

and simplifying:

#d/dx (e^(ex)lnx) = e^(ex)(elnx +1/x)#