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

The answer to (\partial)/(\partial y)(2x^2-3y^2-2x) is -6y

Popular >

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

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

- 6 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (2 x^{2}) = 0

\frac{\partial}{\partial y} (3 y^{2}) = 6 y

\frac{\partial}{\partial y} (2 x) = 0

= 0 - 6 y - 0

Simplify

= - 6 y

\frac{\partial}{\partial x} (sin (x) + cos (- 6 y))

\frac{\partial}{\partial x} (\frac{5 x y ^{2}}{x ^{2} + y ^{4}})

\frac{20}{s \cdot ( s + 10 )}

\int x^{3} - 2.5 x + 1 d x

t \to 7 lim \frac{( t - 7 ) ^{3}}{t - 7}

What is (\partial)/(\partial y)(2x^2-3y^2-2x) ?
The answer to (\partial)/(\partial y)(2x^2-3y^2-2x) is -6y