Anonymous

g (x) = ( sin x ) x^(-2) g ` (x) = ( cos x ) x^(-2) - 2x^(-3) sin x g ` (x) = x^(-3) [ x cos x - 2 sin x ] g ` (x) = [ x cos x - 2 sin x ] / x^3

Anonymous

Quotient rule: d/dx[u(x)/v(x)] = [vu' – uv']/v^2 d/dx[sin(x)/x^2] = [x^2cos(x) – 2xsin(x)]/x^4 = [xcos(x) – 2sin(x)]/x^3

Anonymous

Recall derivative function ratio rule g(x)=f(x)/h(x) ⇒ g’(x)=[f ‘(x)*h(x)-f(x)*h’(x)]/h²(x) In our case f(x)=sinx ⇒ f '(x)=cosx h(x)=x² ⇒ h'(x)=2x hence g'(x)=(cosx*x²-2x*sinx)/x⁴= =(x²cosx -2x*sinx)/x⁴= (xcosx-2sinx)/x³

Anonymous

g(x) = sinx * x^(-2) g'(x) = cos(x)*x^(-2) -2sin(x)*x^(-3)

Anonymous

Adding parentheses to remove ambiguity: g(x)= sin(x)/x^2 g'(x) = [(x²)(sin(x))' - sin(x)(x²)']/(x²)² = [x²cos(x) + 2xsin(x)]/x⁴ = [xcos(x) + 2sin(x)]/x³ = cos(x)/x² + 2sin(x)/x³

Anonymous

rule y = u/v ---> y' = (u'v-uv')/v^2

Anonymous

Try the derivative calculator