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

The answer to (\partial)/(\partial x)(-2ycos^2(x)) is 4ycos(x)sin(x)

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

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

4 y cos (x) sin (x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - 2 y \cdot 2 cos (x) (- sin (x))

Simplify

= 4 y cos (x) sin (x)

\frac{d y}{d x} + y = x^{2} + 1

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

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

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

e^{\frac{1}{6} x^{6}}, 0

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