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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= y^{2} \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)

= y^{2} (\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)

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

Simplify

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

\frac{\partial}{\partial x} (ln (1 + x y))

f (x) = \frac{2 x + 3}{3 x + 2}

\frac{\partial}{\partial x} ((e^{x - y}) g oo g)

f (x) = 1 - 2 cos (x)

\frac{d}{d x} ((\frac{x - 5}{x + 5})^{4})

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