What is the d/(dy)(3x^2y+e^y) ?

The d/(dy)(3x^2y+e^y) is 3x^2+e^y

What is the first d/(dy)(3x^2y+e^y) ?

The first d/(dy)(3x^2y+e^y) is 3x^2+e^y

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\frac{d}{d y} (3 x^{2} y + e^{y})

3 x^{2} + e^{y}

Solution steps

\frac{d}{d y} (3 x^{2} y + e^{y})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (3 x^{2} y) + \frac{d}{d y} (e^{y})

\frac{d}{d y} (3 x^{2} y) = 3 x^{2}

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

= 3 x^{2} + e^{y}

\frac{3 x}{4}

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