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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

x \to - 6 lim (\frac{x + 9}{x + 6})

f (x) = 110 x - 16 x^{2}, at x = 0

\frac{s \cdot ( s + 3 )}{s + 2}

4 t^{3} sin (t)

x y^{^{'}} + y + y^{2} = 0, y (1) = 2

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