What is (\partial)/(\partial y)(yln(y)+ye^x) ?

The answer to (\partial)/(\partial y)(yln(y)+ye^x) is ln(y)+1+e^x

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(\partial)/(\partial y)(yln(y)+ye^x)

\frac{\partial}{\partial y} (y ln (y) + y e^{x})

ln (y) + 1 + e^{x}

Solution steps

\frac{\partial}{\partial y} (y ln (y) + y e^{x})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (y ln (y)) = ln (y) + 1

\frac{\partial}{\partial y} (y e^{x}) = e^{x}

= ln (y) + 1 + e^{x}

\frac{d}{d l} (\frac{d l ^{2} - d - a l ^{3} - b l ^{2}}{l ^{2}})

\int \frac{sin ( 2 x )}{( 1 - cos ^{2} ( x ) ) ^{3}} d x

\frac{- 3}{( s + 3 ) ^{2}}

17 e^{x}

\frac{\partial}{\partial t} (e^{- t})

What is (\partial)/(\partial y)(yln(y)+ye^x) ?
The answer to (\partial)/(\partial y)(yln(y)+ye^x) is ln(y)+1+e^x