What is the solution for derivative of y=ln(e^x) ?

The solution for derivative of y=ln(e^x) is y=(x^2)/2+c_{1}

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Find the Derivative - d/dx y = log of e^x

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

y = \frac{x ^{2}}{2} + c_{1}

Solution steps

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int ln (e^{x}) d x

\int ln (e^{x}) d x = \frac{x ^{2}}{2} + c_{1}

y = \frac{x ^{2}}{2} + c_{1}

sec (x)

sin^{2} (x)

\frac{d}{d x} y = lo g_{2} (x)

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

\int (\frac{( x ^{3} )}{( x ^{4} - 5 )}) d x, u = x^{4} - 5

What is the solution for derivative of y=ln(e^x) ?
The solution for derivative of y=ln(e^x) is y=(x^2)/2+c_{1}