What is (\partial)/(\partial x)(2x^2-8x+y^2-8y+7) ?

The answer to (\partial)/(\partial x)(2x^2-8x+y^2-8y+7) is 4x-8

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(\partial)/(\partial x)(2x^2-8x+y^2-8y+7)

\frac{\partial}{\partial x} (2 x^{2} - 8 x + y^{2} - 8 y + 7)

4 x - 8

Solution steps

\frac{\partial}{\partial x} (2 x^{2} - 8 x + y^{2} - 8 y + 7)

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} (8 x) + \frac{\partial}{\partial x} (y^{2}) - \frac{\partial}{\partial x} (8 y) + \frac{\partial}{\partial x} (7)

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

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

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

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

\frac{\partial}{\partial x} (7) = 0

= 4 x - 8 + 0 - 0 + 0

Simplify

= 4 x - 8

\frac{d}{d x} (\frac{ln ( x ^{3} - y + z )}{x y + z})

\int - 2 x + 2 d x

f (x) = 7 x - ln (x)

y^{^{''}} + y = e^{- t} sin (2 t)

\int 8 θ^{4} d θ

What is (\partial)/(\partial x)(2x^2-8x+y^2-8y+7) ?
The answer to (\partial)/(\partial x)(2x^2-8x+y^2-8y+7) is 4x-8