Q11  x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3

Answers (1)

Given function is 
f(x)=x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3
take u=x ^{x^2 -3}
Now, take log on both the sides
\log u=(x^2-3)\log x
Now, differentiate w.r.t x
\frac{1}{u}.\frac{du}{dx}= \frac{d(x^2-3)}{dx}.\log x+(x^2-3).\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx} = 2x.\log x+(x^2-3).\frac{1}{x}\\ \\ \frac{1}{u}.\frac{du}{dx} = \frac{2x^2\log x+x^2-3}{x}\\ \\ \frac{du}{dx}= u.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\ \\ \frac{du}{dx}= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\ \\                   -(i)
Similarly,
take v=(x-3)^x^2\\
Now, take log on both the sides
\log v=x^2\log (x-3)
Now, differentiate w.r.t x
\frac{1}{v}.\frac{dv}{dx}= \frac{d(x^2)}{dx}.\log (x-3)+x^2.\frac{d(\log (x-3))}{dx}\\ \\ \frac{1}{v}.\frac{dv}{dx} = 2x.\log (x-3)+x^2.\frac{1}{(x-3)}\\ \\ \frac{1}{v}.\frac{dv}{dx} = 2x\log(x-3)+\frac{x^2}{x-3}\\ \\ \frac{dv}{dx}= v.\left ( 2x\log(x-3)+\frac{x^2}{x-3} \right )\\ \\ \frac{dv}{dx}= (x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )\\ \\               -(ii)
Now
f(x)= u + v
f^{'}(x)= \frac{du}{dx}+\frac{dv}{dx}
Put the value from equation (i) and (ii)
f^{'}(x)= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )
Therefore, differentiation w.r.t x is  x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )
 

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