What is (\partial)/(\partial x)(e^{-xy}+y^3x^4) ?

The answer to (\partial)/(\partial x)(e^{-xy}+y^3x^4) is -ye^{-yx}+4y^3x^3

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(\partial)/(\partial x)(e^{-xy}+y^3x^4)

\frac{\partial}{\partial x} (e^{- x y} + y^{3} x^{4})

- y e^{- y x} + 4 y^{3} x^{3}

Solution steps

\frac{\partial}{\partial x} (e^{- x y} + y^{3} x^{4})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (e^{- x y}) + \frac{\partial}{\partial x} (y^{3} x^{4})

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

\frac{\partial}{\partial x} (y^{3} x^{4}) = 4 y^{3} x^{3}

= - y e^{- y x} + 4 y^{3} x^{3}

\frac{d}{d x} (3 x^{3} + 2 x^{2} - 1 x + 5)

\int_{0}^{3} e^{- x^{2}} d x

x \to π lim (ln (cos (x) - 3))

\int y^{2} z + 2 x z^{2} d x

\frac{2}{x}

What is (\partial)/(\partial x)(e^{-xy}+y^3x^4) ?
The answer to (\partial)/(\partial x)(e^{-xy}+y^3x^4) is -ye^{-yx}+4y^3x^3