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

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

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(\partial)/(\partial y)(2xy^4-1)

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

8 x y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 8 x y^{3} - 0

Simplify

= 8 x y^{3}

(5 x^{2} + 11 x)^{\frac{4}{3}}

\int e^{- x (y + 1)} d x

\int_{0}^{50} 9.8 (500 - 10 y) d y

\frac{\partial}{\partial t} (2 t)

f (x) = \frac{6}{4 x + 1}, at x = - 1

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