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

The d/(dy)(e^{x^2+y^2}) is e^{x^2+y^2}*2y

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

The first d/(dy)(e^{x^2+y^2}) is e^{x^2+y^2}*2y

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

e^{x^{2} + y^{2}} \cdot 2 y

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= e^{x^{2} + y^{2}} \cdot 2 y

\frac{d}{d x} (\frac{d y}{d x}) = 4 y

\int 3 csc (x) d x

x^{2} - 9, at x = 2

f^{^{'}} (x) = 3

\frac{d}{d x} (1.1 + 18.9 x^{2})

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