What is (\partial)/(\partial x)(x^2+8xy+17y^2-2y+1) ?

The answer to (\partial)/(\partial x)(x^2+8xy+17y^2-2y+1) is 2x+8y

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(\partial)/(\partial x)(x^2+8xy+17y^2-2y+1)

\frac{\partial}{\partial x} (x^{2} + 8 x y + 17 y^{2} - 2 y + 1)

2 x + 8 y

Solution steps

\frac{\partial}{\partial x} (x^{2} + 8 x y + 17 y^{2} - 2 y + 1)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (8 x y) + \frac{\partial}{\partial x} (17 y^{2}) - \frac{\partial}{\partial x} (2 y) + \frac{\partial}{\partial x} (1)

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

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

\frac{\partial}{\partial x} (17 y^{2}) = 0

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

\frac{\partial}{\partial x} (1) = 0

= 2 x + 8 y + 0 - 0 + 0

Simplify

= 2 x + 8 y

\frac{\partial}{\partial x} (sin (2 t))

\int (- 8 x^{7} y^{- 5} + 6 x^{6} y^{- 4}) d y

\int_{1}^{t} \frac{31}{x ^{3}} d x

x \to 0 lim (\frac{ln ^{2} ( x )}{x})

f (x) = ln^{3} (3 x - 5)

What is (\partial)/(\partial x)(x^2+8xy+17y^2-2y+1) ?
The answer to (\partial)/(\partial x)(x^2+8xy+17y^2-2y+1) is 2x+8y