What is (\partial)/(\partial x)(-2x^3+sin(xy)+4e^{4y}) ?

The answer to (\partial)/(\partial x)(-2x^3+sin(xy)+4e^{4y}) is ycos(xy)-6x^2

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(\partial)/(\partial x)(-2x^3+sin(xy)+4e^{4y})

\frac{\partial}{\partial x} (- 2 x^{3} + sin (x y) + 4 e^{4 y})

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

Solution steps

\frac{\partial}{\partial x} (- 2 x^{3} + sin (x y) + 4 e^{4 y})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (2 x^{3}) + \frac{\partial}{\partial x} (sin (x y)) + \frac{\partial}{\partial x} (4 e^{4 y})

\frac{\partial}{\partial x} (2 x^{3}) = 6 x^{2}

\frac{\partial}{\partial x} (sin (x y)) = cos (x y) y

\frac{\partial}{\partial x} (4 e^{4 y}) = 0

= - 6 x^{2} + cos (x y) y + 0

Simplify

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

x \to 0 lim (∣ x ∣ ln ∣ x ∣)

\frac{1}{s ^{2}} (1 - e^{- s})

\frac{1}{( x ^{2} + 8 ) ^{5}}

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

What is (\partial)/(\partial x)(-2x^3+sin(xy)+4e^{4y}) ?
The answer to (\partial)/(\partial x)(-2x^3+sin(xy)+4e^{4y}) is ycos(xy)-6x^2