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

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

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

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

2 x^{2} y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 2 x^{2} y - 0

Simplify

= 2 x^{2} y

3 + t^{5} + sin (π t)

\int 6 sin^{3} (x) cos^{7} (x) d x

\frac{2 s ^{2} + 5 s + 7}{( s + 1 ) ^{3}}

\frac{d}{d x} (\frac{4 5 x ^{3} + x}{( x - 1 )})

\frac{0.1 \cdot m}{s + 0.1 \cdot m}

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