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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