Question

If $$y=\dfrac{\log x}{x}$$, show that $$\dfrac{d^{2}y}{dx^{2}} =\dfrac{2\log x-3}{x^{3}}$$.

Answer 1

Hint: In this question it is given a function $$y=\dfrac{\log x}{x}$$, we have to find the second order derivative of y, i.e, $$\dfrac{d^{2}y}{dx^{2}}$$.
So to find the solution we have to differentiate y wrt x, so for this we need to know that if a function is in the form of $$y=\dfrac{u}{v}$$ then derivative of ‘y’ is
$$\dfrac{dy}{dx} =\dfrac{v\dfrac{du}{dx} -u\dfrac{dv}{dx} }{v^{2}}$$......(1)
Where u and v are the function of x.

Complete step-by-step answer:
So here the given function is,
$$y=\dfrac{\log x}{x}$$......................(2)
Now differentiating the above equation w.r.t ‘x’ we get,
$$\dfrac{dy}{dx} =\dfrac{d}{dx} \left( \dfrac{\log x}{x} \right) $$...........(3)
So this function is in the form of $$\dfrac{u}{v}$$ where $$u=\log x$$ and $$v=x$$.
So we can apply the formula (1) in the equation (3),
$$\dfrac{dy}{dx} =\dfrac{x\dfrac{d}{dx} \left( \log x\right) -\log x\dfrac{d}{dx} \left( x\right) }{x^{2}}$$
Now as we know that $$\dfrac{d}{dx} \left( \log x\right) =\dfrac{1}{x}$$ and $$\dfrac{d}{dx} \left( x\right) =1$$.
So by using these formulas the above differential equation can be written as,
$$\dfrac{dy}{dx} =\dfrac{x\cdot \dfrac{1}{x} -\log x\cdot 1}{x^{2}}$$
$$\Rightarrow \dfrac{dy}{dx} =\dfrac{1-\log x}{x^{2}}$$............(4)
Now again we are going to differentiate equation (4) w.r.t ‘x’
[where the equation is in the form of $$\dfrac{u}{v}$$ , $$u=1-\log x$$ and $$v=x^{2}$$]
$$\dfrac{d^{2}y}{dx^{2}} =\dfrac{d}{dx} \left( \dfrac{1-\log x}{x^{2}} \right) $$
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{x^{2}\dfrac{d}{dx} \left( 1-\log x\right) -\left( 1-\log x\right) \dfrac{d}{dx} \left( x^{2}\right) }{\left( x^{2}\right)^{2} }$$ [using formula (1)]
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{x^{2}\left( \dfrac{d}{dx} \left( 1\right) -\dfrac{d}{dx} \left( \log x\right) \right) -\left( 1-\log x\right) 2x}{x^{4}}$$ [since, $$\dfrac{d}{dx} \left( x^{n}\right) =nx^{n-1}$$]
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{x^{2}\left( 0-\dfrac{1}{x} \right) -2\left( 1-\log x\right) x}{x^{4}}$$ [since, derivative of constant term is zero]
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{-x^{2}\cdot \dfrac{1}{x} -2x+2x\log x}{x^{4}}$$
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{-x-2x+2x\log x}{x^{4}}$$
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{-3x+2x\log x}{x^{4}}$$
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{x(-3+2\log x)}{x^{4}}$$ [by taking x common from the numerator]
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{(-3+2\log x)}{x^{3}}$$ [canceling x from the numerator and denominator]
$$\Rightarrow \dfrac{d^{2}y}{dx^{2}} =\dfrac{2\log x-3}{x^{3}}$$
Hence this is our required solution.

Note: While solving this type of question you need to know that derivative of any constant term is always zero but when a constant is there with a variable, i.e, for a constant coefficient, then this constant term goes outside of the derivative. Also while derivation you have to keep in mind that if a logarithmic function has ‘e’ in its base then only this function can be differentiable but in the given question they didn’t mentioned about that, so here is another information is that in most of the case when base is ‘e’ then they usually do not mention or do not write in the base.