What is the derivative of 1/x+x ?

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int e^{3 x} (3 + 9 x) d x

3 x + 4

n = 2 \sum \infty \frac{n}{2 ^{n}}

- u

\frac{d y}{d x} y = 3 x^{3} - 8 x^{2} + π

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