What is (\partial)/(\partial x)(2yx^2-3) ?

The answer to (\partial)/(\partial x)(2yx^2-3) is 4yx

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(\partial)/(\partial x)(2yx^2-3)

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

4 y x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (2 y x^{2}) = 4 y x

\frac{\partial}{\partial x} (3) = 0

= 4 y x - 0

Simplify

= 4 y x

f (x) = x^{2} - 3 x + 2

\int_{0}^{1} 7 x e^{- x^{2}} d x

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

\frac{d}{d x} (\frac{6 x ^{4}}{x ^{2} - 3})

f (x) = cos (2 \cdot x) \cdot sin (5 \cdot x)

What is (\partial)/(\partial x)(2yx^2-3) ?
The answer to (\partial)/(\partial x)(2yx^2-3) is 4yx