Question - Finding the Derivative of sin(x)/x^2 using Quotient Rule
Solution:
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 $$x$$.
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