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

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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 ). **

We'll calculate the first derivative of f(x).

We'll put f(x) = y

We'll differentiate with respect to x:

dy/dx =(d/dx) (2x^2 * ln x)

dy/dx =[(d/dx) (2x^2)] * ln x + 2x^2*(d/dx)ln x)

dy/dx = 4xlnx + 2x^2/x

We'll simplify and we'll get:

dy/dx = 4xlnx + 2x

We'll factorize by 2x:

dy/dx = 2x(2lnx + 1)

dy/dx = 2x(ln x^2 + lne)

**dy/dx = 2x ln(e*x^2)**

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**

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