Answer: [tex]\dfrac{1-3ln x}{x^{4}}[/tex]
Step-by-step explanation:
According to Quotient rule in derivatives :
If f(x) and g(x) are differentiable functions , then
[tex]\dfrac{d}{dx}(\dfrac{f(x)}{g(x)})=\dfrac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}[/tex]
The given function : [tex]y=\dfrac{\ln x}{x^3}[/tex]
Differentiate both sides , we get
[tex]\dfrac{dy}{dx}=\dfrac{d}{dx}(\dfrac{\ln x}{x^3})[/tex]
By Quotient rule , this is equal to
[tex]=\dfrac{(\ln x)'(x^3)-(x^3)'(\lnx)}{(x^3)^2}[/tex]
[tex]=\dfrac{(\dfrac{1}{x})(x^3)-\ln x(3x^2)}{x^6}[/tex]
[∵ [tex]\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\ \ \&\ \dfrac{d}{dx}(x^n)=n(x)^{n-1}[/tex]]
[tex]=\dfrac{x^2-3x^2\ln x}{x^6} \\\\=\dfrac{x^2(1-3\ln x)}{x^6}\\\\=\dfrac{(1-3\ln x)}{x^{6-2}}\\\\=\dfrac{1-3\ln x}{x^{4}}[/tex]
Hence, the derivative of the function will be : [tex]\dfrac{1-3ln x}{x^{4}}[/tex]
