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

The answer to (\partial)/(\partial x)(2sin(x^2+y^2)) is 4xcos(x^2+y^2)

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

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

4 x cos (x^{2} + y^{2})

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

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

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

Simplify

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

f (x) = (5)^{x}, at x = 4

\int x^{2} 1 - x^{3} d x

\frac{d}{d θ} (8 sin (θ))

\int 17 x^{3} e^{x} d x

\frac{\partial}{\partial x} (\frac{4 y}{x ^{5}})

What is (\partial)/(\partial x)(2sin(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(2sin(x^2+y^2)) is 4xcos(x^2+y^2)