MCQ
જો $y = {2^{{x^{{2^x}}}}}$ તો $\frac{{dy}}{{dx}} = ...........$
- ✓${2^{{x^{{2^x}}}}}{x^{{2^x}}}{\log _e}2\left( {{2^x}{{\log }_e}x{{\log }_e}2 + \frac{{{2^x}}}{x}} \right)$
- B${2^{{x^{{2^x}}}}}\left( {1 + x{{\log }_e}2} \right){2^x}{\log _e}2$
- C${2^{{x^{{2^x}}}}}\left( {1 + x{{\log }_e}2} \right){2^x}{\left( {{{\log }_e}2} \right)^2}$
- D${2^{{x^{{2^x}}}}}\left( {1 - x{{\log }_e}2} \right){2^x}{\log _e}2$