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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (ln (x)) = \frac{1}{x}

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

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

\int_{- 6}^{8} (\frac{x}{2} + 8) d x

\int \frac{3 t ^{2}}{2} d t

\int x (ln (x))^{2} d x

\frac{d}{d x} (5 ln (2 x + 6))

f (x) = \frac{2 x + 6 x ^{2}}{( 1 + 6 x ) ^{2}}

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