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

The answer to (\partial)/(\partial x)(2x^2+y^2-5y) is 4x

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

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

4 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 4 x + 0 - 0

Simplify

= 4 x

e^{- 2 x} x^{3} (- x + 2)

\int t^{7} e^{- t^{8}} d t

x \to 2 lim (\frac{x}{2 - x})

\frac{d}{d t} (1 - cos (t))

\frac{d}{d x} (cos^{2} (5 - x))

What is (\partial)/(\partial x)(2x^2+y^2-5y) ?
The answer to (\partial)/(\partial x)(2x^2+y^2-5y) is 4x