Q.

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

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a

$x . x^{x - 1} + 7 x^{6} + x 7^{x - 1}$

b

$x^{x} (1 + \log_{e} x) + 7 x^{6} + 7^{x} (\log_{e} 7)$

c

$x^{x} (1 + \log_{e} x) + 7 x^{6} + x . 7^{x - 1}$

d

$x . x^{x - 1} \log_{e} x + 7 x^{6} + 7^{x} (\log_{7} e)$

answer is B.

(Detailed Solution Below)

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Detailed Solution

Given $y = x^{x} + x^{7} + 7^{x} + 7^{7}$

Use the formula $\frac{d}{d x} (x^{x}) = x^{x} (1 + \log x)$ to differentiate the above function

$\begin{array}{rcl} \frac{d y}{d x} & = & x^{x} (1 + \log x) + 7 x^{6} + 7^{x} \log 7 + 0 \\ = & x^{x} (1 + \log x) + 7 x^{6} + 7^{x} \log 7 \end{array}$

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