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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = cos (x^{2} + y^{2})

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

\frac{\partial}{\partial x} (x) = 1

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

= 2 (1 \cdot cos (x^{2} + y^{2}) + (- sin (x^{2} + y^{2}) \cdot 2 x) x)

Simplify

2 (1 \cdot cos (x^{2} + y^{2}) + (- sin (x^{2} + y^{2}) \cdot 2 x) x) : 2 (cos (x^{2} + y^{2}) - 2 x^{2} sin (x^{2} + y^{2}))

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

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

x y + y^{^{'}} = 142 x

\frac{d y}{d x} = 2 x - 1 + 2 x y - y

(- 19.7, 10.5), (- 18.9, 3.4)

\frac{d}{d x} (cos (2 π x))

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