Question

find the differential:y=sinx/x2​

Answer 1

Answer:

(xcosx-2sinx)/x^3

Step-by-step explanation:

Differential of u/v = (u'v-uv')/v^2

So,

Differential of sinx/x^2 = {cosx(x^2) - sinx(2x)} / x^4

(Cancelling one 'x' from each term of numerator and denominator)

(xcosx-2sinx)/x^3

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Answer 2

$\tt y = \frac{sinx}{{x}^{2}} \\ \\ \tt \boxed{\bold{\underline{\green{\tt \frac{d(\frac{u}{v})}{dx} = \frac{\frac{du}{dx}v - \frac{dv}{dx}u}{{v}^{2}} }}}} \\ \\ \tt \therefore \frac{dy}{dx} = \frac{\frac{d(sinx)}{dx}.{x}^{2} - \frac{d({x}^{2})}{dx}.sinx}{{({x}^{2})}^{2}} \\ \\ \tt \therefore \frac{dy}{dx} = \frac{{x}^{2}. cosx - 2x.sinx}{{x}^{4}} \\ \\ \tt \therefore \frac{dy}{dx} = \frac{x(x.cosx - 2sinx)}{x.{x}^{3}} \\ \\ \tt \therefore \frac{dy}{dx} = \frac{\cancel{\orange{x}}(x.cosx - 2sinx)}{\cancel{\orange{x}}.{x}^{3}} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{ \tt \frac{dy}{dx} = \frac{x.cosx - 2sinx}{{x}^{3}} }}}}$