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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= y (- sin (x^{2} y) \cdot 2 y x)

Simplify

y (- sin (x^{2} y) \cdot 2 y x) : - 2 y^{2} x sin (y x^{2})

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

(8 \frac{1}{7}, 9), (4, 3 \frac{1}{5})

y = ln (x^{2} - 5 x + 1), (5, 0)

\int_{0}^{\frac{π}{3}} x sin (6 x) d x

\int (1 + x)^{0} d x

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

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