What is (\partial)/(\partial y)(4x^3y^2-1) ?

The answer to (\partial)/(\partial y)(4x^3y^2-1) is 8x^3y

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(\partial)/(\partial y)(4x^3y^2-1)

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 1)

8 x^{3} y

Solution steps

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (4 x^{3} y^{2}) - \frac{\partial}{\partial y} (1)

\frac{\partial}{\partial y} (4 x^{3} y^{2}) = 8 x^{3} y

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

= 8 x^{3} y - 0

Simplify

= 8 x^{3} y

(y p + 2 y) (p d y + y d p) - (2 y p + 4 y) d y = 0

cos^{2} (13 t)

y = ln (5)

\frac{d y}{d x} = 5 cos^{3} (3 x + 6)

\int \frac{x ^{3}}{x ^{2} - 16} d x

What is (\partial)/(\partial y)(4x^3y^2-1) ?
The answer to (\partial)/(\partial y)(4x^3y^2-1) is 8x^3y