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How do you find the first and second derivative of #ln(x^2e^x)#?

Answer 1

#y' = (2 + x)/x#
#y'' = -2/x^2#

This can be rewritten as

#y = lnx^2 + lne^x#

We know that #y= lnx# and #y = e^x# are inverses, such that #lne = 1#. Using the law #lna^n = nlna#, we can say that #lne^x = x# and that #lnx^2 = 2lnx#.

#y = 2lnx + x#

#y' = 2/x + 1#

This is the first derivative. Differentiate this to get the second derivative.

#y' = 2x^-1 + 1#

#y'' = -2/x^2#

Hopefully this helps!

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Answer 2

William Abdalla

To find the first and second derivatives of ln(x^2e^x), you can use the chain rule and product rule.

First derivative:

Apply the product rule: (f*g)' = f'g + fg'
Differentiate ln(x^2e^x) with respect to x.
First differentiate ln(x^2e^x) using the chain rule: d/dx [ln(u)] = (1/u)*(du/dx)
Then apply the product rule to the remaining terms.

Second derivative:

Differentiate the first derivative obtained above using the same rules.

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Answer from HIX Tutor

When evaluating a one-sided 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.