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How do you find the derivative of #y= e^(ex) ln x#?

Answer 1

Emma Aldridge

#d/dx (e^(ex)lnx) = e^(ex)(elnx +1/x)#

Using the product rule:

(1) #d/dx (e^(ex)lnx) = d/dx ( e^(ex))lnx +e^(ex)d/dx(lnx)#

Now we have using the chain rule:

#d/dx ( e^(ex)) = e^(ex) d/dx (ex) = e*e^(ex)#

while:

#d/dx(lnx) = 1/x#

is a standard derivative.

Substituting this in (1) we have:

#d/dx (e^(ex)lnx) = e*e^(ex)lnx +e^(ex)/x#

and simplifying:

#d/dx (e^(ex)lnx) = e^(ex)(elnx +1/x)#

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Answer 2

Emma Aldridge

To find the derivative of ( y = e^{ex} \ln x ), use the product rule. The derivative is:

[ y' = e^{ex} \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(e^{ex}) ]

[ y' = e^{ex} \cdot \frac{1}{x} + \ln x \cdot e^{ex} \cdot \frac{d}{dx}(ex) ]

[ y' = e^{ex} \cdot \frac{1}{x} + \ln x \cdot e^{ex} \cdot e^x ]

[ y' = e^{ex} \cdot \frac{1}{x} + \ln x \cdot e^{(ex + x)} ]

So, the derivative of ( y = e^{ex} \ln x ) is ( y' = e^{ex} \cdot \frac{1}{x} + \ln x \cdot e^{(ex + x)} ).

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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.