For constant a,frac{d}{d x}left(x^{x}+x^{a}+a | vedclass

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(B) We need to find the derivative of the sum: $\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$.Using the linearity property of the derivative,we h

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$x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$

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(B) We need to find the derivative of the sum: $\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$.Using the linearity property of the derivative,we have:$\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(x^{a}\right)+\frac{d}{d x}\left(a^{x}\right)+\frac{d}{d x}\left(a^{a}\right)$.$1$. For $\frac{d}{d x}(x^x)$: Let $y = x^x$. Taking $\log$ on both sides,$\log y = x \log x$. Differentiating with respect to $x$,$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{dy}{dx} = x^x(1 + \log x)$.$2$. For $\frac{d}{d x}(x^a)$: Using the power rule,$\frac{d}{d x}(x^a) = a x^{a-1}$.$3$. For $\frac{d}{d x}(a^x)$: Using the exponential derivative rule,$\frac{d}{d x}(a^x) = a^x \log a$.$4$. For $\frac{d}{d x}(a^a)$: Since $a$ is a constant,$a^a$ is a constant,so its derivative is $0$.Combining these,the result is $x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$.

For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is

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