What is (\partial)/(\partial x)(3x^2-4y^3) ?

The answer to (\partial)/(\partial x)(3x^2-4y^3) is 6x

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(\partial)/(\partial x)(3x^2-4y^3)

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

6 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (4 y^{3}) = 0

= 6 x - 0

Simplify

= 6 x

\int_{0}^{\frac{π}{2}} cos (3 x) d x

\int (2 e^{x} + 8 x^{3} + 6 x) d x

\int \frac{4}{15 sin ( x ) - 8 cos ( x )} d x

y = 4 - x^{2}

\frac{d y}{d x} = \frac{x ^{2} - 1}{y ^{2} + 1}, y (- 1) = 1

What is (\partial)/(\partial x)(3x^2-4y^3) ?
The answer to (\partial)/(\partial x)(3x^2-4y^3) is 6x