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

The answer to (\partial)/(\partial y)(cos^2(3x-y^2)) is 4ycos(3x-y^2)sin(3x-y^2)

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(\partial)/(\partial y)(cos^2(3x-y^2))

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

= 4 y cos (3 x - y^{2}) sin (3 x - y^{2})

\frac{\partial}{\partial y} (4 x^{0.5} + 2 y^{0.25})

\frac{\partial}{\partial y} (ln (x yz))

\int 7^{x^{2}} d x

x \to \infty lim (\frac{1}{x} + 1)

y^{3} x + y^{2} x^{2} = 6, (2, 1)

What is (\partial)/(\partial y)(cos^2(3x-y^2)) ?
The answer to (\partial)/(\partial y)(cos^2(3x-y^2)) is 4ycos(3x-y^2)sin(3x-y^2)