What is (\partial)/(\partial x)(4cos(x^2+y^2)) ?

The answer to (\partial)/(\partial x)(4cos(x^2+y^2)) is -8xsin(x^2+y^2)

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(\partial)/(\partial x)(4cos(x^2+y^2))

\frac{\partial}{\partial x} (4 cos (x^{2} + y^{2}))

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

Solution steps

\frac{\partial}{\partial x} (4 cos (x^{2} + y^{2}))

Treat

y

as a constant

Take the constant out:

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

= 4 \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

= 4 (- sin (x^{2} + y^{2}) \cdot 2 x)

Simplify

= - 8 x sin (x^{2} + y^{2})

e^{t} cos (6 t)

5 - x^{2}

f (x) = ln (x^{2} + 3)

y = (x + 1) (x + 2)

\int cos (200 t n) d t

What is (\partial)/(\partial x)(4cos(x^2+y^2)) ?
The answer to (\partial)/(\partial x)(4cos(x^2+y^2)) is -8xsin(x^2+y^2)