What is (\partial)/(\partial x)(2x^2ycos(x)) ?

The answer to (\partial)/(\partial x)(2x^2ycos(x)) is 2y(2xcos(x)-x^2sin(x))

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(\partial)/(\partial x)(2x^2ycos(x))

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

2 y (2 x cos (x) - x^{2} sin (x))

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = cos (x)

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

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

\frac{\partial}{\partial x} (cos (x)) = - sin (x)

= 2 y (2 x cos (x) + (- sin (x)) x^{2})

Simplify

= 2 y (2 x cos (x) - x^{2} sin (x))

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

\int (3 x + 3) x^{2} + 2 x + 3 d x

n = 1 \sum \infty 9 - \frac{1}{n ^{2}}

n = 1 \sum \infty 10 (- \frac{2}{3})^{n}

n \to \infty lim (\frac{3}{n})

What is (\partial)/(\partial x)(2x^2ycos(x)) ?
The answer to (\partial)/(\partial x)(2x^2ycos(x)) is 2y(2xcos(x)-x^2sin(x))