What is the derivative of 3x^2-x^2y+y^4 ?

The derivative of 3x^2-x^2y+y^4 is 6x-2xy-x^2(dy)/(dx)+4y^3(dy)/(dx)

What is the first derivative of 3x^2-x^2y+y^4 ?

The first derivative of 3x^2-x^2y+y^4 is 6x-2xy-x^2(dy)/(dx)+4y^3(dy)/(dx)

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

6 x - 2 x y - x^{2} \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

= 6 x - (2 x y + \frac{d y}{d x} x^{2}) + 4 y^{3} \frac{d y}{d x}

Simplify

6 x - (2 x y + \frac{d y}{d x} x^{2}) + 4 y^{3} \frac{d y}{d x} : 6 x - 2 x y - x^{2} \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x}

= 6 x - 2 x y - x^{2} \frac{d y}{d x} + 4 y^{3} \frac{d y}{d x}

\int \frac{x + 7}{x ^{2} + 4 x + 8} d x

6 y^{^{'}} - 3 y = 2 e^{- 2 x}, y (0) = 10

ln (1 + 2 x), 1

\int cos^{3} (\frac{x}{3}) d x

y = x + 2, y = \frac{1}{x + 1}, x = 0, x = 2

What is the derivative of 3x^2-x^2y+y^4 ?
The derivative of 3x^2-x^2y+y^4 is 6x-2xy-x^2(dy)/(dx)+4y^3(dy)/(dx)
What is the first derivative of 3x^2-x^2y+y^4 ?
The first derivative of 3x^2-x^2y+y^4 is 6x-2xy-x^2(dy)/(dx)+4y^3(dy)/(dx)