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

f (x) = x^{3} - 7 x^{2} - 7 x + 4

- 9 t

\int sin^{2} (9 x) d x

\int_{0}^{3} - x + 2 d x

\int 6 x^{5} 3 x^{4} + 2 d x

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