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

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

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

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

2 y^{2} x - 3

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3 x) = 3

= 2 y^{2} x - 3

(- 7, - 4), (- 11, - 2)

y^{^{'}} = \frac{( x ^{2} - y ^{2} )}{x y}, y (2) = 3

y = sin (π - x)

f (x) = 7 x - 17, at x = 3

\frac{d}{d x} (6 x^{\frac{1}{3}} + 3 x^{\frac{4}{3}})

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