What is (\partial)/(\partial x)(14xye^{-x^2-y^2}) ?

The answer to (\partial)/(\partial x)(14xye^{-x^2-y^2}) is 14y(e^{-x^2-y^2}-2e^{-x^2-y^2}x^2)

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(\partial)/(\partial x)(14xye^{-x^2-y^2})

\frac{\partial}{\partial x} (14 x y e^{- x^{2} - y^{2}})

14 y (e^{- x^{2} - y^{2}} - 2 e^{- x^{2} - y^{2}} x^{2})

Solution steps

\frac{\partial}{\partial x} (14 x y e^{- x^{2} - y^{2}})

Treat

y

as a constant

Take the constant out:

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

= 14 y \frac{\partial}{\partial x} (x e^{- x^{2} - y^{2}})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{- x^{2} - y^{2}}

= 14 y (\frac{\partial}{\partial x} (x) e^{- x^{2} - y^{2}} + \frac{\partial}{\partial x} (e^{- x^{2} - y^{2}}) x)

\frac{\partial}{\partial x} (x) = 1

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

= 14 y (1 \cdot e^{- x^{2} - y^{2}} + (- 2 e^{- x^{2} - y^{2}} x) x)

Simplify

14 y (1 \cdot e^{- x^{2} - y^{2}} + (- 2 e^{- x^{2} - y^{2}} x) x) : 14 y (e^{- x^{2} - y^{2}} - 2 e^{- x^{2} - y^{2}} x^{2})

= 14 y (e^{- x^{2} - y^{2}} - 2 e^{- x^{2} - y^{2}} x^{2})

f (x) = x + 3, at x = 6

y^{^{''}} + 16 y = 6 tan (4 x)

\frac{d y}{d x}, x e^{y} - 10 x + 3 y = 0

(sin (3 x))^{l n (x)}

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

What is (\partial)/(\partial x)(14xye^{-x^2-y^2}) ?
The answer to (\partial)/(\partial x)(14xye^{-x^2-y^2}) is 14y(e^{-x^2-y^2}-2e^{-x^2-y^2}x^2)