What is (\partial)/(\partial x)(40xy*e^{5x^2-2y}) ?

The answer to (\partial)/(\partial x)(40xy*e^{5x^2-2y}) is 40y(e^{5x^2-2y}+10e^{5x^2-2y}x^2)

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(\partial)/(\partial x)(40xy*e^{5x^2-2y})

\frac{\partial}{\partial x} (40 x y \cdot e^{5 x^{2} - 2 y})

40 y (e^{5 x^{2} - 2 y} + 10 e^{5 x^{2} - 2 y} x^{2})

Solution steps

\frac{\partial}{\partial x} (40 x y e^{5 x^{2} - 2 y})

Treat

y

as a constant

Take the constant out:

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

= 40 y \frac{\partial}{\partial x} (x e^{5 x^{2} - 2 y})

Apply the Product Rule:

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

f = x, g = e^{5 x^{2} - 2 y}

= 40 y (\frac{\partial}{\partial x} (x) e^{5 x^{2} - 2 y} + \frac{\partial}{\partial x} (e^{5 x^{2} - 2 y}) x)

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

\frac{\partial}{\partial x} (e^{5 x^{2} - 2 y}) = e^{5 x^{2} - 2 y} \cdot 10 x

= 40 y (1 \cdot e^{5 x^{2} - 2 y} + e^{5 x^{2} - 2 y} \cdot 10 xx)

Simplify

40 y (1 \cdot e^{5 x^{2} - 2 y} + e^{5 x^{2} - 2 y} \cdot 10 xx) : 40 y (e^{5 x^{2} - 2 y} + 10 e^{5 x^{2} - 2 y} x^{2})

= 40 y (e^{5 x^{2} - 2 y} + 10 e^{5 x^{2} - 2 y} x^{2})

\int \frac{1}{4} x^{3} d x

\int (\frac{x - x ^{3} e ^{x} + x ^{2}}{x ^{3}}) d x

y^{^{'}} = \frac{1 - y ^{2}}{e ^{x}}

n = 0 \sum \infty p^{n}

x \to 3 lim (6 x)

What is (\partial)/(\partial x)(40xy*e^{5x^2-2y}) ?
The answer to (\partial)/(\partial x)(40xy*e^{5x^2-2y}) is 40y(e^{5x^2-2y}+10e^{5x^2-2y}x^2)