What is the derivative of 2x^4y-4 ?

The derivative of 2x^4y-4 is 2(x^4(dy)/(dx)+4x^3y)

What is the first derivative of 2x^4y-4 ?

The first derivative of 2x^4y-4 is 2(x^4(dy)/(dx)+4x^3y)

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

2 (x^{4} \frac{d y}{d x} + 4 x^{3} y)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 x^{4} y) = 2 (4 x^{3} y + \frac{d y}{d x} x^{4})

\frac{d}{d x} (4) = 0

= 2 (4 x^{3} y + \frac{d y}{d x} x^{4}) - 0

Simplify

= 2 (x^{4} \frac{d y}{d x} + 4 x^{3} y)

y^{^{''''}} - 16 y = 0

\frac{d}{d t} (2 t^{\frac{4}{3}} - 3 t^{\frac{2}{3}})

\frac{d}{d x} (3 e^{x} + \frac{4}{3 x})

e^{x} cos (2 x)

\frac{\partial}{\partial x} (\frac{sin ( x )}{1 + x ^{2}})

What is the derivative of 2x^4y-4 ?
The derivative of 2x^4y-4 is 2(x^4(dy)/(dx)+4x^3y)
What is the first derivative of 2x^4y-4 ?
The first derivative of 2x^4y-4 is 2(x^4(dy)/(dx)+4x^3y)