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

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

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

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

2 x

Solution steps

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

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

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

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

= 2 x + 0 + 0

Simplify

= 2 x

\frac{d ^{2}}{d x ^{2}} (x e^{- x})

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

\int \frac{cos ( x )}{( sin ^{2} ( x ) + 25 ) ^{2}} d x

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

x \frac{d y}{d x} + (- 1 + x) y = - x y^{2}

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