What is (\partial)/(\partial y)(y-cos(y)) ?

The answer to (\partial)/(\partial y)(y-cos(y)) is 1+sin(y)

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(\partial)/(\partial y)(y-cos(y))

\frac{\partial}{\partial y} (y - cos (y))

1 + sin (y)

Solution steps

\frac{\partial}{\partial y} (y - cos (y))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (y) - \frac{\partial}{\partial y} (cos (y))

\frac{\partial}{\partial y} (y) = 1

\frac{\partial}{\partial y} (cos (y)) = - sin (y)

= 1 - (- sin (y))

Simplify

= 1 + sin (y)

\int_{0}^{1} \frac{8}{x ( 1 + x )} d x

f (x) = (x + 2)^{x}

x \to 2 lim (x^{2} + 3 x - 8)

\int 9 + 8 x - x^{2} d x

\int x^{2} e^{11 x} d x

What is (\partial)/(\partial y)(y-cos(y)) ?
The answer to (\partial)/(\partial y)(y-cos(y)) is 1+sin(y)