What is (\partial)/(\partial x)(x^2-x+2y^2+yz-10) ?

The answer to (\partial)/(\partial x)(x^2-x+2y^2+yz-10) is 2x-1

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(\partial)/(\partial x)(x^2-x+2y^2+yz-10)

\frac{\partial}{\partial x} (x^{2} - x + 2 y^{2} + yz - 10)

2 x - 1

Solution steps

\frac{\partial}{\partial x} (x^{2} - x + 2 y^{2} + yz - 10)

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial x} (2 y^{2}) + \frac{\partial}{\partial x} (yz) - \frac{\partial}{\partial x} (10)

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

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

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

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

\frac{\partial}{\partial x} (10) = 0

= 2 x - 1 + 0 + 0 - 0

Simplify

= 2 x - 1

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

y = \frac{x ^{5}}{10} + \frac{1}{6 x ^{3}}

n = 1 \sum \infty \frac{n}{( n + 1 ) !}

\frac{d}{d t} (4 e^{2 t})

\int \frac{e ^{4 x}}{e ^{4 x} + 1} d x

What is (\partial)/(\partial x)(x^2-x+2y^2+yz-10) ?
The answer to (\partial)/(\partial x)(x^2-x+2y^2+yz-10) is 2x-1