What is (\partial)/(\partial x)(8-x^2+xy-4y^2) ?

The answer to (\partial)/(\partial x)(8-x^2+xy-4y^2) is y-2x

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(\partial)/(\partial x)(8-x^2+xy-4y^2)

\frac{\partial}{\partial x} (8 - x^{2} + x y - 4 y^{2})

y - 2 x

Solution steps

\frac{\partial}{\partial x} (8 - x^{2} + x y - 4 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (8) - \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (x y) - \frac{\partial}{\partial x} (4 y^{2})

\frac{\partial}{\partial x} (8) = 0

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

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

\frac{\partial}{\partial x} (4 y^{2}) = 0

= 0 - 2 x + y - 0

Simplify

= y - 2 x

\int \frac{ln ( x )}{e ^{x}} d x

f (x) = x^{2} + 2 x + 1

n = 1 \sum \infty \frac{x ^{n}}{8 ^{n}}

(x^{2} + 1)^{\frac{1}{2}}

What is (\partial)/(\partial x)(8-x^2+xy-4y^2) ?
The answer to (\partial)/(\partial x)(8-x^2+xy-4y^2) is y-2x