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Differentiate w.r.t. \(x\) the function in \( x^{\large x} + x^{\large a} + a^{\large x} + a^{\large a} \), for some fixed \( a > 0 \) and \( x > 0 \)

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  • $\large\frac{d}{dx}$$(x^n)=n(x^{n-1})$
  • $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
Step 1:
Let y=\( x^{\large x} + x^{\large a} + a^{\large x} + a^{\large a} \)
Differentiating with respect to $x$
$\large\frac{dy}{dx}=\frac{d}{dx}$$(x^{\large x})+ax^{\large a-1}+a^{\large x}\log a+0$-----(1)
$a$ is a constant.
Put $u=x^{\large x}$
$\log u=\log x^{\large x}$
$\qquad=x\log x$
$\large\frac{1}{u}\frac{du}{dx}$$=1.\log x+x.\large\frac{1}{x}$
$\qquad\;=\log x+1$
$\large\frac{du}{dx}$$=u(1+\log x)$
Step 2:
Substitute the value of $u$
$\large\frac{du}{dx}=$$x^{\large x}(1+\log x)$
$\large\frac{d}{dx}$$(x^{\large x})$$=x^{\large x}(1+\log x)$
Putting these values in eq(1)
$\large\frac{dy}{dx}=$$x^{\large x}(1+\log x)+ax^{\large a-1}+a^{\large x}\log a$
answered May 14, 2013 by sreemathi.v

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