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  • Differentiate the functions w.r.t. x. log x^x +x ^log x

 Q7 Differentiate the functions w.r.t. x.  (\log x )^x + x ^{\log x }

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best_answer

Given function is
y = (\log x )^x + x ^{\log x }
Let's take t = (\log x)^x
Now, take log on both the sides
\log t = x \log(\log x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}= \log (\log x)+\frac{1}{\log x}\\ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}=(\log x)^x(\log (\log x))+ (\log x )^{x-1}
Similarly, take k = x^{\log x}
Now, take log on both sides 
\log k = \log x \log x = (\log x)^2
Now, differentiate w.r.t. x
We get,
\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x
Therefore, the answer is  (\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x



 

Posted by

Gautam harsolia

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