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What is the derivative of #f(x)=e^(x^2lnx)#?

Answer 1

Aurora Alvarado

#f'(x)=xe^(x^2ln(x))(2ln(x)+1)#

We will use the following:

The chain rule.

The product rule.

#d/dx e^x = e^x#

#d/dx x^n = nx^(n-1)#

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

With those:

#f'(x) = d/dxe^(x^2ln(x))#

#= e^(x^2ln(x))(d/dxx^2ln(x))#

#=e^(x^2ln(x))(x^2(d/dxln(x))+ln(x)(d/dxx^2))#

#=e^(x^2ln(x))(x^2(1/x)+ln(x)(2x))#

#=e^(x^2ln(x))(x+2xln(x))#

#=xe^(x^2ln(x))(2ln(x)+1)#

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Answer 2

Emilia Aguilar

To find the derivative of ( f(x) = e^{x^2 \ln x} ), you can use the chain rule and the product rule. The derivative is:

[ f'(x) = e^{x^2 \ln x} \left(2x \ln x + x + \frac{x}{x \ln x}\right) ]

Simplified:

[ f'(x) = e^{x^2 \ln x} \left(2x \ln x + 1 + \frac{1}{\ln x}\right) ]

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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.