Question & Answer
QUESTION

Find the value of \[{{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\], If \[y=ln{{\left( \dfrac{x}{a+bx} \right)}^{x}}\],
        (a)\[\,{{\left( \dfrac{dy}{dx}+x \right)}^{2}}\]
        (b) \[\,{{\left( \dfrac{dy}{dx}-y \right)}^{2}}\]
       (c) \[{{\left( x\dfrac{dy}{dx}+y \right)}^{2}}\,\]
      (d) \[{{\left( x\dfrac{dy}{dx}-y \right)}^{2}}\]

ANSWER Verified Verified
Hint: To solve this question first apply the log rule to simplify the given function and then proceed with applying the rules of differentiation to get the first derivative again to differentiate the first derivative to get the required answer.

Complete step-by-step solution -
The given expression is \[y=ln{{\left( \dfrac{x}{a+bx} \right)}^{x}}\]
But we know, $\log {{x}^{n}}=n\log x$ , so the given equation becomes,
\[y=x\cdot ln\left( \dfrac{x}{a+bx} \right)\]
Differentiate the given expression with respect to \[x\], we get
 \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( x\cdot ln\left( \dfrac{x}{a+bx} \right) \right)\]
We know that \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\] so the above equation becomes,
\[\dfrac{dy}{dx}=x\dfrac{d}{dx}\left( ln\left( \dfrac{x}{a+bx} \right) \right)+ln\left( \dfrac{x}{a+bx} \right)\dfrac{d}{dx}\left( x \right)\]
We also know that, \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\], so above equation becomes,
\[\dfrac{dy}{dx}=x\times \dfrac{1}{\left( \dfrac{x}{a+bx} \right)}\times \dfrac{d}{dx}\left\{ \dfrac{x}{a+bx} \right\}+\ln \left( \dfrac{x}{a+bx} \right)\times 1\]
Now we will apply the quotient rule, i.e., $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}$ , so the above equation becomes,
\[\dfrac{dy}{dx}=x\times \dfrac{a+bx}{x}\times \left\{ \dfrac{(a+bx)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(a+bx)}{{{\left( a+bx \right)}^{2}}} \right\}+\ln \left( \dfrac{x}{a+bx} \right)\]
We know differentiation of constant term is zero, so the above equation can be written as
\[\dfrac{dy}{dx}=(a+bx)\times \left\{ \dfrac{(a+bx)\times 1-x(b)}{{{\left( a+bx \right)}^{2}}} \right\}+\ln \left( \dfrac{x}{a+bx} \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\left\{ \dfrac{a+bx-bx}{\left( a+bx \right)} \right\}+\ln \left( \dfrac{x}{a+bx} \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\ln \left( \dfrac{x}{a+bx} \right)+\dfrac{a}{a+bx}.........(i)\]
Consider the given expression,
\[y=x\cdot ln\left( \dfrac{x}{a+bx} \right)\]
\[\Rightarrow \dfrac{y}{x}=ln\left( \dfrac{x}{a+bx} \right)...........(ii)\]
Substituting the value from equation (ii) into equation (i), we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{a}{a+bx}..........(iii)\]
Now we will find the second derivative, so differentiating equation (iii) with respect to $'x'$, we get
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{y}{x}+\dfrac{a}{a+bx} \right)\]
Now we will apply the sum rule of differentiation, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{y}{x} \right)+\dfrac{d}{dx}\left( \dfrac{a}{a+bx} \right)\]
Now we will apply the quotient rule, i.e., $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}$ and taking out the constant term, so the above equation becomes,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\dfrac{dy}{dx}-y\dfrac{d(x)}{dx}}{{{x}^{2}}}+a\dfrac{d}{dx}{{\left( a+bx \right)}^{-1}}\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{x\dfrac{dy}{dx}}{{{x}^{2}}}-\dfrac{y}{{{x}^{2}}}+a(-1){{\left( a+bx \right)}^{-1-1}}\dfrac{d}{dx}(a+bx)\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x}\dfrac{dy}{dx}-\dfrac{y}{{{x}^{2}}}-\dfrac{ab}{{{\left( a+bx \right)}^{2}}}\]
Now substitute value from equation (iii), we get

\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x}\left( \dfrac{y}{x}+\dfrac{a}{a+bx} \right)-\dfrac{y}{{{x}^{2}}}-\dfrac{ab}{{{\left( a+bx \right)}^{2}}}\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{y}{{{x}^{2}}}+\dfrac{a}{x\left( a+bx \right)}-\dfrac{y}{{{x}^{2}}}-\dfrac{ab}{{{\left( a+bx \right)}^{2}}}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{a}{x\left( a+bx \right)}-\dfrac{ab}{{{\left( a+bx \right)}^{2}}}\]
Taking the LCM and solving, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{a(a+bx)-xab}{x{{\left( a+bx \right)}^{2}}}\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{a}^{2}}+abx-xab}{x{{\left( a+bx \right)}^{2}}}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{a}^{2}}}{x{{\left( a+bx \right)}^{2}}}=\dfrac{1}{x}{{\left( \dfrac{a}{a+bx} \right)}^{2}}.........(iv)\]
Now consider equation (iii),
\[\dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{a}{a+bx}\]
\[\therefore \dfrac{a}{a+bx}=\dfrac{dy}{dx}-\dfrac{y}{x}\]
Put this value in equation (iv), we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x}{{\left( \dfrac{dy}{dx}-\dfrac{y}{x} \right)}^{2}}\]
Multiply both sides by \[{{x}^{3}}\], we get
\[\Rightarrow {{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{x}^{3}}}{x}{{\left( \dfrac{dy}{dx}-\dfrac{y}{x} \right)}^{2}}\]
\[\Rightarrow {{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{x}^{2}}{{\left( \dfrac{dy}{dx}-\dfrac{y}{x} \right)}^{2}}\]
\[\Rightarrow {{x}^{3}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( x\dfrac{dy}{dx}-y \right)}^{2}}\]
Hence the correct answer is option (d).

 Note: We can also use the formula \[\dfrac{d}{dx}\left\{ f{{\left( x \right)}^{g\left( x \right)}} \right\}=f{{\left( x \right)}^{g\left( x \right)}}\times \dfrac{d}{dx}\left\{ g(x)\ln \left( f(x) \right) \right\}\]
In this method also we will get the same result.