What is (\partial)/(\partial x)(2x^2-16x+y^2+1) ?

The answer to (\partial)/(\partial x)(2x^2-16x+y^2+1) is 4x-16

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(\partial)/(\partial x)(2x^2-16x+y^2+1)

\frac{\partial}{\partial x} (2 x^{2} - 16 x + y^{2} + 1)

4 x - 16

Solution steps

\frac{\partial}{\partial x} (2 x^{2} - 16 x + y^{2} + 1)

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} (16 x) + \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (1)

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

\frac{\partial}{\partial x} (16 x) = 16

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

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

= 4 x - 16 + 0 + 0

Simplify

= 4 x - 16

\frac{\partial}{\partial x} (e^{t} x + 1)

\int x^{2} e^{x - 3} d x

x \to 0 lim (sin (x) - cos (x))

\int \frac{x ^{4} - x ^{3} - x - 1}{x ^{3} - x ^{2}} d x

What is (\partial)/(\partial x)(2x^2-16x+y^2+1) ?
The answer to (\partial)/(\partial x)(2x^2-16x+y^2+1) is 4x-16