# How do you find the first and second derivative of #x^2 lnx#?

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To find the first derivative of (x^2 \ln(x)), you can use the product rule, which states that if you have two functions (u(x)) and (v(x)), then the derivative of their product (u(x)v(x)) with respect to (x) is (u'(x)v(x) + u(x)v'(x)).

Applying the product rule, the first derivative of (x^2 \ln(x)) is:

[ \frac{d}{dx} (x^2 \ln(x)) = 2x\ln(x) + x ]

To find the second derivative, you differentiate the first derivative obtained above with respect to (x).

Using the product rule again, the second derivative of (x^2 \ln(x)) is:

[ \frac{d^2}{dx^2} (x^2 \ln(x)) = \frac{d}{dx} (2x\ln(x) + x) = 2\ln(x) + \frac{2}{x} + 1 ]

So, the first derivative is (2x\ln(x) + x), and the second derivative is (2\ln(x) + \frac{2}{x} + 1).

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