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

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

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

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

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

Solution steps

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

Simplify

cos (- (2 y^{2} + 4 x)) : cos (2 y^{2} + 4 x)

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

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

\int (x^{3} + 3 x^{2}) d x

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