What is (\partial)/(\partial x)(3x^2y^2-1) ?

The answer to (\partial)/(\partial x)(3x^2y^2-1) is 6xy^2

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(\partial)/(\partial x)(3x^2y^2-1)

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 1)

6 x y^{2}

Solution steps

\frac{\partial}{\partial x} (3 x^{2} y^{2} - 1)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x^{2} y^{2}) - \frac{\partial}{\partial x} (1)

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

\frac{\partial}{\partial x} (1) = 0

= 6 y^{2} x - 0

Simplify

= 6 x y^{2}

x^{2} - 1

n = 2 \sum \infty \frac{1}{n ( n + 1 )}

\int x^{\frac{- 2}{3}} x^{\frac{1}{3}} - 2 d x

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

\frac{d}{d x} (\frac{2}{x} - \frac{3 x}{3})

What is (\partial)/(\partial x)(3x^2y^2-1) ?
The answer to (\partial)/(\partial x)(3x^2y^2-1) is 6xy^2