What is (\partial)/(\partial x)(2x^2+y^2-4y+4) ?

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

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(\partial)/(\partial x)(2x^2+y^2-4y+4)

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

4 x

Solution steps

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

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} (4 y) + \frac{\partial}{\partial x} (4)

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

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

\frac{\partial}{\partial x} (4 y) = 0

\frac{\partial}{\partial x} (4) = 0

= 4 x + 0 - 0 + 0

Simplify

= 4 x

0.5 (e^{x} + e^{- x}), 0 \leq x \leq ln (2)

\frac{d}{d x} (\frac{lo g _{10} ( x )}{ln ( x )} + 2 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{x}

\frac{d}{d x} (in (sec (x)))

n = 1 \sum \infty \frac{2}{e ^{n - 3}}

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