What is (\partial)/(\partial x)(cos(5x-4y^2)) ?

The answer to (\partial)/(\partial x)(cos(5x-4y^2)) is -sin(5x-4y^2)*5

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(\partial)/(\partial x)(cos(5x-4y^2))

\frac{\partial}{\partial x} (cos (5 x - 4 y^{2}))

- sin (5 x - 4 y^{2}) \cdot 5

Solution steps

\frac{\partial}{\partial x} (cos (5 x - 4 y^{2}))

Treat

y

as a constant

Apply the chain rule:

- sin (5 x - 4 y^{2}) \frac{\partial}{\partial x} (5 x - 4 y^{2})

= - sin (5 x - 4 y^{2}) \frac{\partial}{\partial x} (5 x - 4 y^{2})

\frac{\partial}{\partial x} (5 x - 4 y^{2}) = 5

= - sin (5 x - 4 y^{2}) \cdot 5

\frac{d}{d x} (\frac{x ^{4} + 6}{x ^{2}})

y^{^{'}} = \frac{t ^{2} + 3 y ^{2}}{2 t y}

\frac{\partial}{\partial x} (x^{8} y^{5} + 4 x^{6} y)

\frac{\partial}{\partial y} (2 x^{4} y - 3)

\int x (x - 5)^{5} d x

What is (\partial)/(\partial x)(cos(5x-4y^2)) ?
The answer to (\partial)/(\partial x)(cos(5x-4y^2)) is -sin(5x-4y^2)*5