What is the derivative of xln(x^2+x) ?

The derivative of xln(x^2+x) is ln(x^2+x)+(2x+1)/(x+1)

What is the first derivative of xln(x^2+x) ?

The first derivative of xln(x^2+x) is ln(x^2+x)+(2x+1)/(x+1)

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\frac{d}{d x} (x ln (x^{2} + x))

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

Solution steps

\frac{d}{d x} (x ln (x^{2} + x))

Apply the Product Rule:

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

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

\int \frac{e ^{2 t}}{2} d t

x \to \frac{π}{2} + lim (cos (x))

\int (\frac{4}{x} - 1) d x

\frac{\partial}{\partial x} (x sin (4 x^{2} y))

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

What is the derivative of xln(x^2+x) ?
The derivative of xln(x^2+x) is ln(x^2+x)+(2x+1)/(x+1)
What is the first derivative of xln(x^2+x) ?
The first derivative of xln(x^2+x) is ln(x^2+x)+(2x+1)/(x+1)