Given #y = e^((ln x)^2)# how do you find find y'(e)?
2 Answers
May 2, 2017
# dy/dx = (2e^(ln^2x)lnx)/x => y'(e) = 2#
Explanation:
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
# dy/dx = (2e^(ln^2e)lne)/e #
# " " = (2e^1*1)/e #
# " " = 2 #
May 2, 2017
Recall that
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#