What is (\partial)/(\partial x)(ln(x^2+x)) ?

The answer to (\partial)/(\partial x)(ln(x^2+x)) is (2x+1)/(x^2+x)

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(\partial)/(\partial x)(ln(x^2+x))

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{85 t}{99 t - 85}

x \to 3 lim (x^{2} + 2 x - 1)

v + x \cdot \frac{d v}{d x} = x^{2} - 1 + 2 v

b cos (x) + x^{2} sin (x)

What is (\partial)/(\partial x)(ln(x^2+x)) ?
The answer to (\partial)/(\partial x)(ln(x^2+x)) is (2x+1)/(x^2+x)