What is (\partial)/(\partial x)(cos(2(2yx-1))) ?

The answer to (\partial)/(\partial x)(cos(2(2yx-1))) is -sin(2(2yx-1))*4y

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(\partial)/(\partial x)(cos(2(2yx-1)))

\frac{\partial}{\partial x} (cos (2 (2 y x - 1)))

- sin (2 (2 y x - 1)) \cdot 4 y

Solution steps

\frac{\partial}{\partial x} (cos (2 (2 y x - 1)))

Treat

y

as a constant

Apply the chain rule:

- sin (2 (2 y x - 1)) \frac{\partial}{\partial x} (2 (2 y x - 1))

= - sin (2 (2 y x - 1)) \frac{\partial}{\partial x} (2 (2 y x - 1))

\frac{\partial}{\partial x} (2 (2 y x - 1)) = 4 y

= - sin (2 (2 y x - 1)) \cdot 4 y

\frac{\partial}{\partial x} (e^{x yz - 2})

x \to 0 + lim (\frac{2}{x ^{\frac{1}{3}}})

sin^{2} (π x)

\frac{\partial}{\partial x} (98 y^{2} + x^{2} - x^{2} y)

\frac{d}{d x} ((x - 4)^{2} - 5)

What is (\partial)/(\partial x)(cos(2(2yx-1))) ?
The answer to (\partial)/(\partial x)(cos(2(2yx-1))) is -sin(2(2yx-1))*4y