What is (\partial)/(\partial y)(e^{y^2}-e^y) ?

The answer to (\partial)/(\partial y)(e^{y^2}-e^y) is e^{y^2}*2y-e^y

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(\partial)/(\partial y)(e^{y^2}-e^y)

\frac{\partial}{\partial y} (e^{y^{2}} - e^{y})

e^{y^{2}} \cdot 2 y - e^{y}

Solution steps

\frac{\partial}{\partial y} (e^{y^{2}} - e^{y})

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (e^{y^{2}}) - \frac{\partial}{\partial y} (e^{y})

\frac{\partial}{\partial y} (e^{y^{2}}) = e^{y^{2}} \cdot 2 y

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

= e^{y^{2}} \cdot 2 y - e^{y}

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What is (\partial)/(\partial y)(e^{y^2}-e^y) ?
The answer to (\partial)/(\partial y)(e^{y^2}-e^y) is e^{y^2}*2y-e^y