Home > Calculus > Product Rule

How do you differentiate # f(x) = x² ln(x) #?

Answer 1

Christopher Agresta

#f'(x)=2xlnx+x#

Apply the product rule: if

#f(x)=g(x)h(x)#, then

#f'(x)=g'(x)h(x)+g(x)h'(x)#

So,

#g(x)=x^2# #h(x)=lnx#

and

#g'(x)=2x# #h'(x)=1/x#

To summarize everything,

#f'(x)=2x(lnx)+x^2(1/x)#

#=>2xlnx+x#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

To differentiate the function ( f(x) = x^2 \ln(x) ), you can use the product rule of differentiation. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), their product's derivative is ( u'(x)v(x) + u(x)v'(x) ). Applying this rule:

[ f'(x) = (x^2)' \ln(x) + x^2 (\ln(x))' ]

To find the derivative of each term:

( (x^2)' = 2x )
( (\ln(x))' = \frac{1}{x} )

Substituting these derivatives back into the expression:

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

[ f'(x) = 2x \ln(x) + x ]

So, the derivative of ( f(x) = x^2 \ln(x) ) is ( f'(x) = 2x \ln(x) + x ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.