What is (\partial)/(\partial x)(2ye^{x^2+y^2}) ?

The answer to (\partial)/(\partial x)(2ye^{x^2+y^2}) is 4e^{x^2+y^2}yx

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(\partial)/(\partial x)(2ye^{x^2+y^2})

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

= 4 e^{x^{2} + y^{2}} y x

\frac{d}{d x} (ln (1 + \frac{5}{x}))

f (x) = x^{2} + 11

t \to \frac{π}{3} lim (cos (t) + 2)

t \to 0 lim (\frac{2 t}{tan ( t )})

y^{^{'}} = 6 e^{(2 x - y)}, y (0) = 0

What is (\partial)/(\partial x)(2ye^{x^2+y^2}) ?
The answer to (\partial)/(\partial x)(2ye^{x^2+y^2}) is 4e^{x^2+y^2}yx