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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

- sin (4 y^{2} - 2 x) \frac{\partial}{\partial x} (4 y^{2} - 2 x)

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

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

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

Simplify

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

\frac{d}{d x} (sin (x^{t a n (x)}))

(2 x + 1) y^{^{'}} + 2 y = 4 x, y (0) = 1

\frac{\partial}{\partial z} (ln (x^{2} + y^{4}) - z)

25 y^{^{''}} - y = x e^{\frac{x}{5}}, y (0) = 1, y^{^{'}} (0) = 0

x \to \infty lim (\frac{1}{x ( ln ( x ) ) ^{4}})

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