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

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

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

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

cos (x^{2} + y^{2}) \cdot 2 x

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

\int cos (15 x) d x

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

x \to - 5 - lim (3)

y^{^{'}} = 6 - y

x \to 3 - lim (\frac{x ^{2}}{9 - x ^{2}})

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