What is derivative of y(θ)=arctan(cos(θ)) ?

The answer to derivative of y(θ)=arctan(cos(θ)) is (dy)/(dx)=(-sin(x)-y(cos^2(x)+1))/(x(cos^2(x)+1))

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derivative of y(θ)=arctan(cos(θ))

derivative of

y = arctan (cos (x))

\frac{d y}{d x} = \frac{- sin ( x ) - y ( cos ^{2} ( x ) + 1 )}{x ( cos ^{2} ( x ) + 1 )}

Solution steps

y x = arctan (cos (x))

Treat

y

y (x)

Differentiate both sides:

x \frac{d y}{d x} + y = - \frac{sin ( x )}{cos ^{2} ( x ) + 1}

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- sin ( x ) - y ( cos ^{2} ( x ) + 1 )}{x ( cos ^{2} ( x ) + 1 )}

\frac{d y}{d x} = \frac{- sin ( x ) - y ( cos ^{2} ( x ) + 1 )}{x ( cos ^{2} ( x ) + 1 )}

\int x (2 - x)^{2} d x

\int_{0}^{5} x 1 + x d x

e^{x^{3} + 2 x}

\int \frac{7 + x + x ^{2}}{x} d x

f (x) = 8 x^{2} - x

What is derivative of y(θ)=arctan(cos(θ)) ?
The answer to derivative of y(θ)=arctan(cos(θ)) is (dy)/(dx)=(-sin(x)-y(cos^2(x)+1))/(x(cos^2(x)+1))