What is (\partial)/(\partial y)(-20xy^2+6x) ?

The answer to (\partial)/(\partial y)(-20xy^2+6x) is -40xy

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(\partial)/(\partial y)(-20xy^2+6x)

\frac{\partial}{\partial y} (- 20 x y^{2} + 6 x)

- 40 x y

Solution steps

\frac{\partial}{\partial y} (- 20 x y^{2} + 6 x)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial y} (20 x y^{2}) + \frac{\partial}{\partial y} (6 x)

\frac{\partial}{\partial y} (20 x y^{2}) = 40 x y

\frac{\partial}{\partial y} (6 x) = 0

= - 40 x y + 0

Simplify

= - 40 x y

x y^{^{'}} = y^{2}, y (1) = 1

t \cdot y^{^{'}} - 3 \cdot y = t^{4}

(t + 2)^{2} e^{t}

t \cdot e^{- t}

f (x) = x^{3} e^{x}

What is (\partial)/(\partial y)(-20xy^2+6x) ?
The answer to (\partial)/(\partial y)(-20xy^2+6x) is -40xy