What is the derivative of 1/x+x^2 ?

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

What is the first derivative of 1/x+x^2 ?

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

f (x) = \frac{x}{x - 3}

f (x) = \frac{x ^{2} + x - 2}{x ^{3} + 6}

\frac{d}{d x} (2 x 4 - x^{2})

\frac{d}{d x} (ln (e^{x} - e^{a}))

\frac{d}{d x} (e^{x^{2} + 9 x})

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