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

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

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

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

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derivative of

ln (e^{x} + x)

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x + 6

\frac{d}{d x} (\frac{x ^{2} - 3 x - 4}{x})

4 x y^{^{'}} + y = 20 x

\int sec^{4} (3 x) tan (3 x) d x

\int \frac{( x ^{2} - 2 x + 5 )}{x ^{2}} d x

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