1

I am doing it by $x^2$ as the and differentiating both of the above separately and then deciding them I got answer $x^{x^2} (2\log x+1))/2$

I don't know if it is correct

gt6989b
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IRSAD HUSSAIN
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3 Answers3

1

Perhaps it may be easier by substituting $u = x^2 \iff x = \pm \sqrt{u}$. Depending on the context of your problem, you may know which branch of the $\sqrt{x}$ to take, let's assume $x = \sqrt{u}$. Then you need to find $$ \frac{d}{du} \left[ \left(u^{1/2}\right)^u\right] = \frac{d}{du} \left[ u^{u/2}\right] = \frac{d}{du} \exp\left(\ln \left( u^{u/2} \right)\right) = \frac{d}{du} \exp\left(\frac{u}{2} \ln u\right) $$ which you can compute by chain rule and back-substitute.

gt6989b
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0

$$\frac{dx^{x^2}}{dx^2}=\frac{de^{x^2\log x}}{2x\,dx}=\frac{2x\log x+x}{2x}x^{x^2}=\frac{2\log x+1}2x^{x^2}.$$

Alternatively,

$$\frac{d\sqrt{x^2}^{x^2}}{dx^2}=\frac{de^{{x^2\log x^2/2}}}{dx^2}=\frac{\log x^2+1}2e^{{x^2\log x^2/2}}=\frac{2\log x+1}2x^{x^2}.$$

0

$$2x\frac{dx^{x^2}}{dx^2}=\frac{dx^2}{dx}\frac{dx^{x^2}}{dx^2}=\frac{dx^{x^2}}{dx}\tag1$$

Making use of $x^{x^2}=e^{x^2\ln x}$ for the RHS of $(1)$ we find:$$\cdots=x^{x^2}[2x\ln x+x]$$ So dividing both sides with $2x$ we arrive at:$$\frac{dx^{x^2}}{dx^2}=\frac12x^{x^2}[2\ln x+1]$$

drhab
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