MCQ
If $y = {(\tan x)^{\cot x}}$, then ${{dy} \over {dx}} =$
- ✓$y\cos {\rm{e}}{{\rm{c}}^2}x(1 - \log \tan x)$
- B$y\,{\rm{cos}}{\rm{e}}{{\rm{c}}^2}x(1 + \log \tan x)$
- C$y\cos {\rm{e}}{{\rm{c}}^2}x(\log \tan x)$
- DNone of these
==> $\frac{1}{y}\frac{{dy}}{{dx}} = {\rm{cose}}{{\rm{c}}^{\rm{2}}}x - \log \tan x.{\rm{cose}}{{\rm{c}}^2}x$
==> $\frac{{dy}}{{dx}} = y{\rm{cose}}{{\rm{c}}^2}x(1 - \log \tan x)$.
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