What is derivative of y(θ)=θsin(θ) ?

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

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derivative of y(θ)=θsin(θ)

derivative of

y = x sin (x)

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

Solution steps

y x = x sin (x)

Treat

y

y (x)

Differentiate both sides:

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

Isolate

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

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

{\frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4}}

y^{^{'}} = 2 x y, y (1) = 1

\frac{d}{d x} (\frac{5}{sin ( x ) + cos ( x )})

\int_{2}^{x} \frac{( 2 x - 3 )}{6} d x

x \to - 5 lim (sin (x \cdot π))

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