Given #y = e^((ln x)^2)# how do you find find y'(e)?
So:
Then:
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# dy/dx = (2e^(ln^2x)lnx)/x => y'(e) = 2#
We have:
Take Natural Logarithms of both sides:
Differentiate Implicitly, and apply the chain rule:
Which we can rearrange to get:
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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:

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} ]

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

Simplify: [ y'(\text{e}) = e^{(\ln \text{e})^2} \cdot 2 \cdot \frac{1}{\text{e}} ]

Remember that ( \ln \text{e} = 1 ): [ y'(\text{e}) = e^{1^2} \cdot 2 \cdot \frac{1}{\text{e}} ]

Simplify further: [ y'(\text{e}) = e^1 \cdot 2 \cdot \frac{1}{\text{e}} = 2 ]
So, ( y'(\text{e}) = 2 ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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