If y=xx+x7+7x+77, then dydx=

y=xx+x7+7x+77 differentiate w.r.to x on both sides ⇒dydx=ddx(xx)+ddx(x7)+ddx(7x)+ddx(77) ⇒dydx=xx1+logx+7x6+7x log7+0. ⇒dydx=xx1+logx+7x6+7x log7

If y=xx+x7+7x+77, then dydx=

y=xx+x7+7x+77 differentiate w.r.to x on both sides ⇒dydx=ddx(xx)+ddx(x7)+ddx(7x)+ddx(77) ⇒dydx=xx1+logx+7x6+7x log7+0. ⇒dydx=xx1+logx+7x6+7x log7

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

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

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

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

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

answer is B.

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Detailed Solution

$y = x^{x} + x^{7} + 7^{x} + 7^{7} d i f f e r e n t i a t e w . r . t o x o n b o t h s i d e s \Rightarrow \frac{d y}{d x} = \frac{d}{d x} (x^{x}) + \frac{d}{d x} (x^{7}) + \frac{d}{d x} (7^{x}) + \frac{d}{d x} (7^{7}) \Rightarrow \frac{dy}{dx} = x^{x} (1 + logx) + 7 x^{6} + 7^{x} \log 7 + 0 . \Rightarrow \frac{dy}{dx} = x^{x} (1 + logx) + 7 x^{6} + 7^{x} \log 7$