Given #y = e^((ln x)^2)# how do you find find y'(e)?

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Answer 1 · 2017-05-02T20:50:09

# dy/dx = (2e^(ln^2x)lnx)/x => y'(e) = 2#

We have:

# y = e^(ln^2x) #

Take Natural Logarithms of both sides:

# ln y = (lnx)^2 #

Differentiate Implicitly, and apply the chain rule:

# 1/y * dy/dx = 2(lnx)*(1/x) #

Which we can rearrange to get:

# dy/dx = (2ylnx)/x #
# " " = (2e^(ln^2x)lnx)/x #

So then, when #x=e# we have:

# dy/dx = (2e^(ln^2e)lne)/e #
# " " = (2e^1*1)/e #
# " " = 2 #

Answer 2 · 2017-05-02T22:33:07

Recall that #d/dxe^x=e^x#. Then, through the chain rule, #d/dxe^f(x)=e^f(x)f'(x)#.

So:

#y(x)=e^((lnx)^2)#

#y'(x)=e^((lnx)^2)d/dx(lnx)^2#

#color(white)(y'(x))=e^((lnx)^2)(2(lnx))d/dxlnx#

#color(white)(y'(x))=(2e^((lnx)^2)lnx)/x#

Then:

#y'(e)=(2e^((lne)^2)lne)/e#

#color(white)(y'(e))=(2e^1(1))/e#

#color(white)(y'(e))=2#