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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

y^{^{''}} - 4 y^{^{'}} = 6 t

\frac{7 s ^{3} - 9 s + 1}{3 s ^{5}}

\int \frac{3}{x ( x ^{4} + 1 )} d x

\frac{\partial}{\partial z} (cos (x y))

x y y^{^{'}} - y^{2} = x^{4}

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