# Differentiate f(x) = 2x^2 * ln x

## Expert Answers We have to differentiate f(x) = 2x^2 * ln x

f(x) = 2x^2 * ln x

Now we use the product rule to find the derivative. The derivative of f(x)*g(x) = f'(x)*g(x) + f(x)*g'(x).

=> f'(x) = 2*[x^2* [d(ln x) /dx] + [d(x^2)/dx]*ln x]

=>f'(x) = 2[x^2 * (1/x) + 2x* ln x]

=> f'(x) = 2x^2/x + 4x*ln x

=> f'(x) = 2x + 4x*ln x

Therefore the required expression is 2x + 4x*ln x

Approved by eNotes Editorial Team Given the function f(x) = 2x^2 * ln x.

We need to determine the first derivative.

We notice that the function f(x) is a product of two functions.

Then, we will use the product rule to find the derivative.

Let f(x) = u*v such that:

==> u= 2x^2  ==>   u' = 4x

==> v = ln(x)   ==>    v' = 1/x

Then we know that:

f'(x) = u'*v + u*v'

Let us substitute.

==> f'(x) = 4x*lnx + 2x^2 * 1/x

==> f'(x) = 4x*lnx + 2x

==> We will factor 2x from both terms.

==> f'(x) = 2x*( 1 + 2lnx ).

Approved by eNotes Editorial Team

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