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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int sin (2 x) cos (7 x) d x

\int 5 tan^{5} (x) sec^{4} (x) d x

\int_{6}^{8} \frac{12}{( x - 6 ) ^{3}} d x

y = 6 x^{2} - 2 x, y = x^{3} - 6 x + 2

n \to + (- \infty) + lim (e^{n})

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