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

The answer to (\partial)/(\partial x)(ycos(x^2+y^2)) is -2yxsin(x^2+y^2)

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(\partial)/(\partial x)(ycos(x^2+y^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int \frac{1100 x}{x ^{2} + 80} d x

f (x) = x, (64, 8)

\frac{\partial}{\partial x} (cos (x - y))

What is (\partial)/(\partial x)(ycos(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(ycos(x^2+y^2)) is -2yxsin(x^2+y^2)