What is (\partial)/(\partial x)(2y^2-2xy+7x-5) ?

The answer to (\partial)/(\partial x)(2y^2-2xy+7x-5) is -2y+7

Popular >

(\partial)/(\partial x)(2y^2-2xy+7x-5)

\frac{\partial}{\partial x} (2 y^{2} - 2 x y + 7 x - 5)

- 2 y + 7

Solution steps

\frac{\partial}{\partial x} (2 y^{2} - 2 x y + 7 x - 5)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 y^{2}) - \frac{\partial}{\partial x} (2 x y) + \frac{\partial}{\partial x} (7 x) - \frac{\partial}{\partial x} (5)

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

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

\frac{\partial}{\partial x} (7 x) = 7

\frac{\partial}{\partial x} (5) = 0

= 0 - 2 y + 7 - 0

Simplify

= - 2 y + 7

\int (x \cdot sec^{2} (x)) d x

\int x^{7} + 7 x + \frac{7}{x} + 7 d x

\int e^{x^{3} + 5} x^{2} d x

\int (x - 6) x + 7 d x

\frac{d}{d x} (x^{2} - 4 x + 7)

What is (\partial)/(\partial x)(2y^2-2xy+7x-5) ?
The answer to (\partial)/(\partial x)(2y^2-2xy+7x-5) is -2y+7