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

The answer to (\partial)/(\partial x)(2x+yz-2) is 2

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(\partial)/(\partial x)(2x+yz-2)

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

2

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 2 + 0 - 0

Simplify

= 2

\int_{0}^{1} 1 - 4 x d x

\int \frac{16 + x ^{2}}{2 x ^{4}} d x

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

d er i v a t i v e f (x) = ln (cos (x))

\frac{d}{d x} (ln (x^{2} + 2))

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