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 Q10    x ^x + x ^a + a ^x + a ^a  , for some fixed a > 0 and x > 0 

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Given function is
f(x)=x ^x + x ^a + a ^x + a ^a
Lets take 
u = x^x
Now, take log on both sides
\log u = x \log x
Now, differentiate w.r.t x
\frac{1}{u}.\frac{du}{dx}= \frac{dx}{dx}.\log x+x.\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx}= 1.\log x+x.\frac{1}{x}\\ \\ \frac{du}{dx}= y.(\log x+1)\\ \\ \frac{du}{dx}= x^x.(\log x+1)           -(i)
Similarly, take v = x^a
take log on both the sides
\log v = a\log x
Now, differentiate w.r.t x
\frac{1}{v}.\frac{dv}{dx}= a.\frac{d(\log x)}{dx}=a.\frac{1}{x}= \frac{a}{x}\\ \\ \frac{dv}{dx}= v.\frac{a}{x}\\ \\ \frac{dv}{dx}= x^a.\frac{a}{x}           -(ii)

Similarly, take z = a^x
take log on both the sides
\log z = x\log a
Now, differentiate w.r.t x
\frac{1}{z}.\frac{dz}{dx}=\log a.\frac{d(x)}{dx}=\log a.1= \log a\\ \\ \frac{dz}{dx}= z.\log a\\ \\ \frac{dz}{dx}= a^x.\log a    -(iii)

Similarly, take w = a^a
take log on both the sides
\log w = a\log a= \ constant
Now, differentiate w.r.t x
\frac{1}{w}.\frac{dw}{dx}= a.\frac{d(a\log a)}{dx}= 0\\ \\ \frac{dw}{dx} = 0                          -(iv)
Now,
f(x)=u+v+z+w
f^{'}(x) = \frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}
Put values from equation (i) , (ii) ,(iii) and (iv)
f^{'}(x)= x^x(\log x+1)+ax^{a-1}+a^x\log a
Therefore, differentiation w.r.t. x is  x^x(\log x+1)+ax^{a-1}+a^x\log a

Posted by

Gautam harsolia

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