Example Question - derivative of sin x

Finding the Derivative of sin(x)/x^2 using Quotient Rule

To find the derivative \(\frac{dy}{dx}\) of the given function \(y = \frac{\sin x}{x^2}\), you can use the quotient rule. The quotient rule states that if you have a function in the form of \(\frac{f(x)}{g(x)}\), then its derivative is given by \[ \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \] Let \(f(x) = \sin x\) and \(g(x) = x^2\). We need to find the derivatives \(f'(x)\) and \(g'(x)\). \(f'(x) = \cos x\) (since the derivative of \(\sin x\) with respect to \(x\) is \(\cos x\)) \(g'(x) = 2x\) (since the derivative of \(x^2\) with respect to \(x\) is \(2x\)) Now apply the quotient rule: \[ \frac{dy}{dx} = \frac{(\cos x)(x^2) - (\sin x)(2x)}{(x^2)^2} \] Simplifying the expression: \[ \frac{dy}{dx} = \frac{x^2\cos x - 2x\sin x}{x^4} \] This derivative can further be simplified if necessary by factoring out an \(x\) in the numerator (assuming \(x \neq 0\) since we cannot divide by zero): \[ \frac{dy}{dx} = \frac{x(\cos x - 2\frac{\sin x}{x})}{x^4} \] This simplifies to: \[ \frac{dy}{dx} = \frac{\cos x}{x^3} - 2\frac{\sin x}{x^4} \] So this is the derivative of the function \(y = \frac{\sin x}{x^2}\) with respect to \(x\).