What is (\partial)/(\partial x)(4ycos(xy)) ?

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

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(\partial)/(\partial x)(4ycos(xy))

\frac{\partial}{\partial x} (4 y cos (x y))

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

Solution steps

\frac{\partial}{\partial x} (4 y cos (x y))

Treat

y

as a constant

Take the constant out:

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

= 4 y \frac{\partial}{\partial x} (cos (x y))

Apply the chain rule:

- sin (x y) \frac{\partial}{\partial x} (x y)

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

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

= 4 y (- sin (x y) y)

Simplify

4 y (- sin (x y) y) : - 4 y^{2} sin (y x)

= - 4 y^{2} sin (y x)

x tan (x)

\int_{\frac{π}{2}}^{π} sin^{2} (x) d x

5 - 4 x^{3}

{y}

\int \frac{3 x ^{2} 16 + y ^{2}}{y}

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