What is the first (x^2ln(x))^' ?

The first (x^2ln(x))^' is 2xln(x)+x

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(x^2ln(x))^'

(x^{2} ln (x))^{^{'}}

2 x ln (x) + x

Solution steps

(x^{2} ln (x))^{'}

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = ln (x)

= (x^{2})^{^{'}} ln (x) + (ln (x))^{^{'}} x^{2}

(x^{2})^{'} = 2 x

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

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

\frac{1}{x} x^{2} = x

= 2 x ln (x) + x

x \to 0 + lim (x e^{\frac{2}{x}})

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