What is (\partial)/(\partial y)(4cos(x^2+y^2)) ?

The answer to (\partial)/(\partial y)(4cos(x^2+y^2)) is -8ysin(x^2+y^2)

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(\partial)/(\partial y)(4cos(x^2+y^2))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

= - 8 y sin (x^{2} + y^{2})

x \to 0 lim (\frac{ln ( 1 + 5 x )}{x})

y^{^{'}} - \frac{y}{x} = 0

\int (4 x + 3) \cdot (2 x^{2} + 3 x)^{3} d x

\int \frac{( x ^{3} )}{x ^{2} + 64} d x

\frac{d}{d x} ((3 x + 2) (x - 1)^{3})

What is (\partial)/(\partial y)(4cos(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(4cos(x^2+y^2)) is -8ysin(x^2+y^2)