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

The answer to derivative of y(θ)=cos(θ)sin(θ) is (dy}{dx}=\frac{cos(2x)-y)/x

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derivative of y(θ)=cos(θ)sin(θ)

derivative of

y = cos (x) sin (x)

\frac{d y}{d x} = \frac{cos ( 2 x ) - y}{x}

Solution steps

y x = cos (x) sin (x)

Treat

y

y (x)

Differentiate both sides:

x \frac{d y}{d x} + y = cos (2 x)

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{cos ( 2 x ) - y}{x}

\frac{d y}{d x} = \frac{cos ( 2 x ) - y}{x}

\int \frac{sin ( 2 x )}{38 + cos ^{2} ( x )} d x

\frac{\partial}{\partial z} (9 x^{\frac{y}{z}})

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

x \to 2 lim (3^{x} + ln (x - 1))

f (2) = \frac{4}{1 + x ^{2}}, at x = 1

What is derivative of y(θ)=cos(θ)sin(θ) ?
The answer to derivative of y(θ)=cos(θ)sin(θ) is (dy}{dx}=\frac{cos(2x)-y)/x