What is (\partial)/(\partial x)(sin^2(3xy^2-y)) ?

The answer to (\partial)/(\partial x)(sin^2(3xy^2-y)) is 6y^2sin(3xy^2-y)cos(3xy^2-y)

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(\partial)/(\partial x)(sin^2(3xy^2-y))

\frac{\partial}{\partial x} (sin^{2} (3 x y^{2} - y))

6 y^{2} sin (3 x y^{2} - y) cos (3 x y^{2} - y)

Solution steps

\frac{\partial}{\partial x} (sin^{2} (3 x y^{2} - y))

Treat

y

as a constant

Apply the chain rule:

2 sin (3 x y^{2} - y) \frac{\partial}{\partial x} (sin (3 x y^{2} - y))

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

\frac{\partial}{\partial x} (sin (3 x y^{2} - y)) = cos (3 x y^{2} - y) \cdot 3 y^{2}

= 2 sin (3 x y^{2} - y) cos (3 x y^{2} - y) \cdot 3 y^{2}

Simplify

= 6 y^{2} sin (3 x y^{2} - y) cos (3 x y^{2} - y)

\frac{d y}{d x}, y = 1 1^{x}

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

\int \frac{3 x ^{2} + 1}{x ^{5} - x} d x

(0.5) (- 2.4)

y^{^{'}} = 5 y + y^{6}, y (0) = 1

What is (\partial)/(\partial x)(sin^2(3xy^2-y)) ?
The answer to (\partial)/(\partial x)(sin^2(3xy^2-y)) is 6y^2sin(3xy^2-y)cos(3xy^2-y)