If y=x^x+x^7+7^x+7^7,then frac{dy}{dx}= | vedclass

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(B) Given $y = x^x + x^7 + 7^x + 7^7$. To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$ separately. $1$. For $x^x$: Let $u = x^x

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$x^x(1+\log_e x)+7 x^6+7^x(\log_e 7)$

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(B) Given $y = x^x + x^7 + 7^x + 7^7$. To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$ separately. $1$. For $x^x$: Let $u = x^x$. Taking $\log$ on both sides,$\log u = x \log x$. Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{du}{dx} = x^x(1 + \log_e x)$. $2$. For $x^7$: Using the power rule,$\frac{d}{dx}(x^7) = 7x^6$. $3$. For $7^x$: Using the exponential rule $\frac{d}{dx}(a^x) = a^x \log_e a$,we get $\frac{d}{dx}(7^x) = 7^x \log_e 7$. $4$. For $7^7$: Since $7^7$ is a constant,its derivative is $0$. Combining these,$\frac{dy}{dx} = x^x(1 + \log_e x) + 7x^6 + 7^x \log_e 7 + 0$. Therefore,$\frac{dy}{dx} = x^x(1 + \log_e x) + 7x^6 + 7^x \log_e 7$.

If $y=x^x+x^7+7^x+7^7$,then $\frac{dy}{dx}=$

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