What is (\partial)/(\partial x)(8x^2+y^2-9y) ?

The answer to (\partial)/(\partial x)(8x^2+y^2-9y) is 16x

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(\partial)/(\partial x)(8x^2+y^2-9y)

\frac{\partial}{\partial x} (8 x^{2} + y^{2} - 9 y)

16 x

Solution steps

\frac{\partial}{\partial x} (8 x^{2} + y^{2} - 9 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (8 x^{2}) + \frac{\partial}{\partial x} (y^{2}) - \frac{\partial}{\partial x} (9 y)

\frac{\partial}{\partial x} (8 x^{2}) = 16 x

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

\frac{\partial}{\partial x} (9 y) = 0

= 16 x + 0 - 0

Simplify

= 16 x

f (x) = x^{2} + 3 x + 1, at x = 4

x \to 0 lim (ln (0))

\int x^{3} (3 + x^{4})^{5} d x

\frac{\partial}{\partial x} (2 x^{2} y + 3)

\frac{x + 1}{x - 1}

What is (\partial)/(\partial x)(8x^2+y^2-9y) ?
The answer to (\partial)/(\partial x)(8x^2+y^2-9y) is 16x