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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int y e^{- y^{2}} d y

y = - 3 - 7 x^{2}

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

\frac{\partial}{\partial x} (2 e^{x^{2}})

x \to 0 lim (\frac{14}{x})

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