What is (\partial)/(\partial x)(6xye^{-x^2-y^2}) ?

The answer to (\partial)/(\partial x)(6xye^{-x^2-y^2}) is 6y(e^{-x^2-y^2}-2e^{-x^2-y^2}x^2)

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(\partial)/(\partial x)(6xye^{-x^2-y^2})

\frac{\partial}{\partial x} (6 x y e^{- x^{2} - y^{2}})

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

Solution steps

\frac{\partial}{\partial x} (6 x y e^{- x^{2} - y^{2}})

Treat

y

as a constant

Take the constant out:

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

= 6 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}}

= 6 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

= 6 y (1 \cdot e^{- x^{2} - y^{2}} + (- 2 e^{- x^{2} - y^{2}} x) x)

Simplify

6 y (1 \cdot e^{- x^{2} - y^{2}} + (- 2 e^{- x^{2} - y^{2}} x) x) : 6 y (e^{- x^{2} - y^{2}} - 2 e^{- x^{2} - y^{2}} x^{2})

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

y = 6 sec (x), at x = \frac{π}{6}

\int cos^{4} (d) x d x

x^{2} y^{^{''}} + 3 x y^{^{'}} + 3 y = 0

\frac{\partial}{\partial t} (4 t)

f (x) = 3 x + 8

What is (\partial)/(\partial x)(6xye^{-x^2-y^2}) ?
The answer to (\partial)/(\partial x)(6xye^{-x^2-y^2}) is 6y(e^{-x^2-y^2}-2e^{-x^2-y^2}x^2)