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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

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

\int (80 - 3 t - (2 t + 30)) d t

\frac{d}{d x} (\frac{x - x ^{2}}{2 - 3 x + x ^{2}})

\int_{0}^{l n (6)} e^{x} - e^{- 3 x} d x

y = \frac{4 x}{9 - tan ( x )}

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