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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} ((x - 1)^{4})

\int sin^{3} (θ) cos^{8} (θ) d θ

\int \frac{3}{3 sin ( x ) - 4 cos ( x )} d x

\frac{1}{cot ( \frac{1}{t} )}

\frac{d}{d x} (3^{c o s (2 x)})

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