What is (\partial)/(\partial x)(x^2y^3+6xyz-2yz) ?

The answer to (\partial)/(\partial x)(x^2y^3+6xyz-2yz) is 2xy^3+6yz

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(\partial)/(\partial x)(x^2y^3+6xyz-2yz)

\frac{\partial}{\partial x} (x^{2} y^{3} + 6 x yz - 2 yz)

2 x y^{3} + 6 yz

Solution steps

\frac{\partial}{\partial x} (x^{2} y^{3} + 6 x yz - 2 yz)

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2} y^{3}) + \frac{\partial}{\partial x} (6 x yz) - \frac{\partial}{\partial x} (2 yz)

\frac{\partial}{\partial x} (x^{2} y^{3}) = 2 y^{3} x

\frac{\partial}{\partial x} (6 x yz) = 6 yz

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

= 2 y^{3} x + 6 yz - 0

Simplify

= 2 x y^{3} + 6 yz

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

\frac{\partial}{\partial x} (7 x e^{x^{2}} + y^{2})

\int \frac{23}{( x ^{2} - 1 ) ^{2}} d x

n = 1 \sum \infty \frac{n - 1}{5 n - 1}

(x^{2} + 2 y^{^{'}}) y^{^{''}} + 2 x y^{^{'}} = 0

What is (\partial)/(\partial x)(x^2y^3+6xyz-2yz) ?
The answer to (\partial)/(\partial x)(x^2y^3+6xyz-2yz) is 2xy^3+6yz