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

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

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

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

6 x y

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 6 y x + 0 + 0

Simplify

= 6 x y

\frac{d}{d x} (\frac{6 x}{x ^{2} - 3})

x \to - 5 lim (\frac{x ^{3} + 125}{x ^{2} - 25})

\int \frac{x - 2}{x + 7} d x

\int_{0}^{9} x^{2} - 6 x + 5 d x

x \to 1 - lim (\frac{∣ x - 1 ∣}{1 - x})

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