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

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

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

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

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

Solution steps

\frac{\partial}{\partial x} (2 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} (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

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

Simplify

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

\frac{d}{d x} (x + 13)

x \to \infty lim (ln ∣ x ∣)

y^{^{''}} + 81 y = 0, y (0) = 4, y^{^{'}} (0) = 7

im pl i c i t \frac{d y}{d x}, \frac{y}{x} = 0

x \to 0 lim (ln (x) sin (x))

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