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

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

Popular >

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

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

2 x + 1

Solution steps

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

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

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

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

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

= 2 x - 0 + 1 + 0

Simplify

= 2 x + 1

\frac{2 x ^{2} + 4 x + 6}{x}

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

- 4 \frac{d y}{d x} + 12 x^{2} = 0

y = 2 x - 3

y = x + 6, y = x^{3}, x = 0

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