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

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

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

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

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

Solution steps

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

Treat

y

as a constant

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

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

\int (2400 x) 3 + 2 x d x

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

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

\int \frac{x ^{3}}{x ^{2} - 36} d x

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