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

Question

Elijah Abram · Accepted Answer

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#

Ezekiel Alexander · Answer

# 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 #

Isabella Ackerman · Answer

undefined To find $ y'(\text{e}) $ for the function $ y = e^{(\ln x)^2} $, you'll need to use the chain rule.

Given that $ y = e^{(\ln x)^2} $, take the derivative of $ y $ with respect to $ x $, and then evaluate it at $ x = \text{e} $.

Here's the process:

1. Find $ y' $ using the chain rule:
$$ y' = \frac{d}{dx} \left( e^{(\ln x)^2} \right) = e^{(\ln x)^2} \cdot 2(\ln x) \cdot \frac{1}{x} $$

2. Substitute $ x = \text{e} $ into $ y' $:
$$ y'(\text{e}) = e^{(\ln \text{e})^2} \cdot 2(\ln \text{e}) \cdot \frac{1}{\text{e}} $$

3. Simplify:
$$ y'(\text{e}) = e^{(\ln \text{e})^2} \cdot 2 \cdot \frac{1}{\text{e}} $$

4. Remember that $ \ln \text{e} = 1 $:
$$ y'(\text{e}) = e^{1^2} \cdot 2 \cdot \frac{1}{\text{e}} $$

5. Simplify further:
$$ y'(\text{e}) = e^1 \cdot 2 \cdot \frac{1}{\text{e}} = 2 $$

So, $ y'(\text{e}) = 2 $.