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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\int e^{- \frac{y}{x}} d y

\int \frac{sin ( x )}{1 - sin ^{2} ( x )} d x

f (x) = \frac{1 - cos ( x )}{x}

\int_{0}^{4} \frac{x}{2} d x

(1 - x^{2}) y^{^{'}} - x y = 1

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