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

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

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(\partial)/(\partial x)(2yx^2-cos(y))

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

4 y x

Solution steps

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

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} (cos (y))

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

\frac{\partial}{\partial x} (cos (y)) = 0

= 4 y x - 0

Simplify

= 4 y x

\int \frac{1}{x ^{2} 225 - x ^{2}} d x

\frac{d y}{d x} = sin (6 x)

y^{^{'}} = 3 - 2 y

\int 7 tan^{2} (x) d x

6 e^{5 t} cos (2 t) - e^{7 t}

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