What is (\partial ^2)/(\partial y^2)(2cos(xy^2)) ?

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

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(\partial ^2)/(\partial y^2)(2cos(xy^2))

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

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

Solution steps

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

\frac{\partial}{\partial y} (2 cos (x y^{2})) = - 4 x y sin (x y^{2})

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

\frac{\partial}{\partial y} (- 4 x y sin (x y^{2})) = - 4 x (sin (x y^{2}) + 2 x y^{2} cos (x y^{2}))

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

f (x) = 4 x^{2} - 5 x, at x = - 3

\frac{\partial}{\partial x} (x^{2} ln (y - 1))

\int_{1}^{2} x (x - 1)^{4} d x

\int ln (t) \cdot t^{2} d t

x \to + (- \infty) + lim (x^{5})

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