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

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

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

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

1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 1 + 0 - 0

Simplify

= 1

\int (t^{2} + 1)^{2} d t

\int_{- \infty}^{9} r e^{\frac{r}{3}} d r

\int \frac{4}{x 1 + 2 x} d x

\frac{\partial}{\partial x} (x \cdot e^{- \frac{y}{x}})

n = 1 \sum \infty \frac{( x ) ^{n - 1}}{x}

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