What is (\partial)/(\partial x)(10xye^{-(x^2+y^2)}) ?

The answer to (\partial)/(\partial x)(10xye^{-(x^2+y^2)}) is 10y(-2e^{-x^2-y^2}x^2+e^{-x^2-y^2})

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(\partial)/(\partial x)(10xye^{-(x^2+y^2)})

\frac{\partial}{\partial x} (10 x y e^{- (x^{2} + y^{2})})

10 y (- 2 e^{- x^{2} - y^{2}} x^{2} + e^{- x^{2} - y^{2}})

Solution steps

\frac{\partial}{\partial x} (10 x y e^{- (x^{2} + y^{2})})

Simplify

10 x y e^{- (x^{2} + y^{2})} : 10 e^{- x^{2} - y^{2}} x y

= \frac{\partial}{\partial x} (10 e^{- x^{2} - y^{2}} x y)

Treat

y

as a constant

Take the constant out:

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

= 10 y \frac{\partial}{\partial x} (e^{- x^{2} - y^{2}} x)

Apply the Product Rule:

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

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

= 10 y (\frac{\partial}{\partial x} (e^{- x^{2} - y^{2}}) x + \frac{\partial}{\partial x} (x) e^{- x^{2} - y^{2}})

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

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

= 10 y ((- 2 e^{- x^{2} - y^{2}} x) x + 1 \cdot e^{- x^{2} - y^{2}})

Simplify

10 y ((- 2 e^{- x^{2} - y^{2}} x) x + 1 \cdot e^{- x^{2} - y^{2}}) : 10 y (- 2 e^{- x^{2} - y^{2}} x^{2} + e^{- x^{2} - y^{2}})

= 10 y (- 2 e^{- x^{2} - y^{2}} x^{2} + e^{- x^{2} - y^{2}})

\int_{- 1}^{1} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

f (x) = tan (4 x)

4

- 4 cos (t)

n = 0 \sum \infty \frac{3 n}{4 n - 1}

What is (\partial)/(\partial x)(10xye^{-(x^2+y^2)}) ?
The answer to (\partial)/(\partial x)(10xye^{-(x^2+y^2)}) is 10y(-2e^{-x^2-y^2}x^2+e^{-x^2-y^2})