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

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

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

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

- 4 sin (x + 2 y)

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= 2 (- sin (x + 2 y) \cdot 2)

Simplify

= - 4 sin (x + 2 y)

\int \frac{x ^{2} - 4}{x ( x ^{2} + 4 )} d x

\int cos^{5} (θ) sin (θ) d θ

y = x, y = 2 x 49 - x^{2}

y^{^{''}} + 25 y = 0, y (π) = 3, y^{^{'}} (π) = - 3

\int_{0}^{4} (- x^{2} + 4 x) d x

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