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Question: If \[y = {\left( {x\log x} \right)^{\log (\log x)}}\], then \( \dfrac{{dy}}{{dx}} = \) 1) \( {\le...

If y=(xlog⁡x)log⁡(log⁡x)y = {\left( {x\log x} \right)^{\log (\log x)}}, then dydx=\dfrac{{dy}}{{dx}} =

  1. {\left( {x\log x} \right)^{\log \left( {\log x} \right)}}\left\\{ {\left( {\dfrac{{\log x + \log \left( {\log x} \right)}}{{x\log x}}} \right) + \left( {\log \left( {\log x} \right)} \right)\left[ {\dfrac{1}{x} + \dfrac{1}{{x\log x}}} \right]} \right\\}
  2. \left\\{ {{{\left( {x\log x} \right)}^{\log \left( {\log x} \right)}}\log \left( {\log x} \right) + \left[ {\dfrac{2}{{\log x}} + \dfrac{1}{x}} \right]} \right\\}
  3. (xlog⁡x)xlog⁡x[log⁡(log⁡x)x][1log⁡x+1]{\left( {x\log x} \right)^{x\log x}}\left[ {\dfrac{{\log \left( {\log x} \right)}}{x}} \right]\left[ {\dfrac{1}{{\log x}} + 1} \right]
  4. [ylog⁡yxlog⁡x][2log⁡(log⁡x)+1]\left[ {\dfrac{{y\log y}}{{x\log x}}} \right]\left[ {2\log \left( {\log x} \right) + 1} \right]
Explanation

Solution

Hint : To solve this problem, we will first take logarithmic function on both sides, as for instance, given, y=f(x)y = f\left( x \right) , then taking logarithmic function on both sides will give, log⁡y=log⁡(f(x))\log y = \log \left( {f\left( x \right)} \right) . Then we will differentiate the terms accordingly. And, in the end, we will change the sides of all the terms to the right hand side and will be left with dydx\dfrac{{dy}}{{dx}} on the left hand side.

Complete step-by-step answer :
Given, y=(xlog⁡x)log⁡(log⁡x)y = {\left( {x\log x} \right)^{\log (\log x)}}
Now, taking logarithmic function on both sides, we get,
\log y = \log \left\\{ {{{\left( {x\log x} \right)}^{\log (\log x)}}} \right\\}
We know, log⁡ab=blog⁡a\log {a^b} = b\log a .
Using this property in above equation, we get,
\Rightarrow \log y = \log \left( {\log x} \right).\log \left\\{ {x\log x} \right\\}
Also, we know, log⁡(ab)=log⁡a+log⁡b\log \left( {ab} \right) = \log a + \log b .
Using this property, we get,
\Rightarrow \log y = \log \left( {\log x} \right).\left\\{ {\log x + \log \left( {\log x} \right)} \right\\}
Now, differentiating both sides with respect to xx , we get,
⇒1ydydx=ddx[log⁡(log⁡x).log⁡x+log⁡(log⁡x)]\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {\log \left( {\log x} \right).\\{ \log x + \log \left( {\log x} \right)\\} } \right]
We know the product rule of differentiation ddx(uv)=dudx.v+u.dvdx\dfrac{d}{{dx}}\left( {uv} \right) = \dfrac{{du}}{{dx}}.v + u.\dfrac{{dv}}{{dx}} .
Using this property in above equation, we get,
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left\\{ {\log \left( {\log x} \right)} \right\\} \times \left[ {\log x + \log \left( {\log x} \right)} \right] + \log \left( {\log x} \right) \times \dfrac{d}{{dx}}\left\\{ {\log x + \log \left( {\log x} \right)} \right\\}
Using the chain rule of differentiation d(f[g(x)])dx=f′[g(x)]×g′(x)\dfrac{{d\left( {f\left[ {g\left( x \right)} \right]} \right)}}{{dx}} = f'\left[ {g\left( x \right)} \right] \times g'\left( x \right), we get,
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \times \dfrac{d}{{dx}}\left( {\log x} \right)\left\\{ {\log x + \log \left( {\log x} \right)} \right\\} + \log \left( {\log x} \right)\left\\{ {\dfrac{1}{x} + \dfrac{1}{{\log x}}\dfrac{d}{{dx}}\left( {\log x} \right)} \right\\}
We know the derivative of the logarithmic function is 1x\dfrac{1}{x} . So, we get,
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\log x}} \times \dfrac{1}{x} \times \left\\{ {\log x + \log \left( {\log x} \right)} \right\\} + \log \left( {\log x} \right)\left\\{ {\dfrac{1}{x} + \dfrac{1}{{\log x}} \times \dfrac{1}{x}} \right\\}
\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{{\left\\{ {\log x + \log \left( {\log x} \right)} \right\\}}}{{x\log x}} + \log \left( {\log x} \right)\left\\{ {\dfrac{1}{x} + \dfrac{1}{{x\log x}}} \right\\}
Now, multiplying both sides by yy , we get,
\Rightarrow \dfrac{{dy}}{{dx}} = y\left[ {\dfrac{{\left\\{ {\log x + \log \left( {\log x} \right)} \right\\}}}{{x\log x}} + \log \left( {\log x} \right)\left\\{ {\dfrac{1}{x} + \dfrac{1}{{x\log x}}} \right\\}} \right]
Now, substituting the original value of yy , we get,
\Rightarrow \dfrac{{dy}}{{dx}} = {\left( {x\log x} \right)^{\log \left( {\log x} \right)}}\left[ {\dfrac{{\left\\{ {\log x + \log \left( {\log x} \right)} \right\\}}}{{x\log x}} + \log \left( {\log x} \right)\left\\{ {\dfrac{1}{x} + \dfrac{1}{{x\log x}}} \right\\}} \right]
So, the correct answer is “Option 1”.

Note : Sometimes the basic chain rule can’t be applied to a function, so we have to operate the function to change it to a form that is easily differentiable. We may have to do that by using logarithmic functions like in this question to solve it. Sometimes we may have to take the logarithmic function more than once to solve the problem. In case of some infinite function, we have to take the infinite part as the dependent variable and then differentiate accordingly. For example,
y=x+x+x+x.....y = \sqrt {x + \sqrt {x + \sqrt {x + \sqrt {x.....} } } }
It is an infinite function. We can rewrite it as,
y=x+yy = \sqrt {x + y}
Now, we can differentiate accordingly.