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

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

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

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

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

2 y + e^{x}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d y} (2 e^{x}) = 0

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

= e^{x} + 0 + 2 y

Simplify

= 2 y + e^{x}

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\int 40 x^{3} + 20 d x

\int 20 - (20)^{2} - x^{2} d x

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