What is (\partial)/(\partial x)(-2e^{4y-x^2-y^2}x) ?

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

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(\partial)/(\partial x)(-2e^{4y-x^2-y^2}x)

\frac{\partial}{\partial x} (- 2 e^{4 y - x^{2} - y^{2}} x)

- 2 (- 2 e^{4 y - x^{2} - y^{2}} x^{2} + e^{4 y - x^{2} - y^{2}})

Solution steps

\frac{\partial}{\partial x} (- 2 e^{4 y - x^{2} - y^{2}} x)

Treat

y

as a constant

Take the constant out:

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

= - 2 \frac{\partial}{\partial x} (e^{4 y - x^{2} - y^{2}} x)

Apply the Product Rule:

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

f = e^{4 y - x^{2} - y^{2}}, g = x

= - 2 (\frac{\partial}{\partial x} (e^{4 y - x^{2} - y^{2}}) x + \frac{\partial}{\partial x} (x) e^{4 y - x^{2} - y^{2}})

\frac{\partial}{\partial x} (e^{4 y - x^{2} - y^{2}}) = - 2 e^{4 y - x^{2} - y^{2}} x

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

= - 2 ((- 2 e^{4 y - x^{2} - y^{2}} x) x + 1 \cdot e^{4 y - x^{2} - y^{2}})

Simplify

- 2 ((- 2 e^{4 y - x^{2} - y^{2}} x) x + 1 \cdot e^{4 y - x^{2} - y^{2}}) : - 2 (- 2 e^{4 y - x^{2} - y^{2}} x^{2} + e^{4 y - x^{2} - y^{2}})

= - 2 (- 2 e^{4 y - x^{2} - y^{2}} x^{2} + e^{4 y - x^{2} - y^{2}})

x \to \infty lim (x - 6)

(8 cos (π x)), (8 x^{2} - 2), [- 0.5, 0.5]

\frac{\partial}{\partial y} ((x - 4) ln (x y))

θ \to 0 lim (sin (θ))

\frac{1}{2 r} + \frac{1}{8 r ^{\frac{7}{8}}}

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