How do you find the derivative of #y=e^x*lnx#?
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To find the derivative of ( y = e^x \cdot \ln(x) ), you can use the product rule. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their product, ( u(x) \cdot v(x) ), is given by the formula:
[ (u \cdot v)' = u'v + uv' ]
For ( y = e^x \cdot \ln(x) ), let ( u(x) = e^x ) and ( v(x) = \ln(x) ).
Now, we find the derivatives of ( u(x) ) and ( v(x) ):
[ u'(x) = e^x ] [ v'(x) = \frac{1}{x} ]
Applying the product rule:
[ y' = (e^x \cdot \ln(x))' = u'v + uv' = e^x \cdot \frac{1}{x} + e^x \cdot \ln(x) ]
So, the derivative of ( y = e^x \cdot \ln(x) ) is:
[ y' = e^x \cdot \frac{1}{x} + e^x \cdot \ln(x) ]
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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.
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.
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.
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.

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