If y = (x ^3 x)^ 3/2 , then dy dx = - Vidyadip

MCQ

If $y = {(x{\cot ^3}x)^{3/2}},$ then ${{dy} \over {dx}} = $

✓
${3 \over 2}{(x{\cot ^3}x)^{1/2}}[{\cot ^3}x - 3x{\cot ^2}x\cos {\rm{e}}{{\rm{c}}^2}x]$
B
${3 \over 2}{(x{\cot ^3}x)^{1/2}}[{\cot ^2}x - 3x{\cot ^2}x\,{\rm{cose}}{{\rm{c}}^2}x]$
C
${3 \over 2}{(x{\cot ^3}x)^{1/3}}[{\cot ^3}x - 3x\,{\rm{cos}}{\rm{e}}{{\rm{c}}^2}x]$
D
${3 \over 2}{(x{\cot ^3}x)^{3/2}}[{\cot ^3}x - 3x\,{\rm{cos}}{\rm{e}}{{\rm{c}}^2}x]$

✓

Answer

Correct option: A.

${3 \over 2}{(x{\cot ^3}x)^{1/2}}[{\cot ^3}x - 3x{\cot ^2}x\cos {\rm{e}}{{\rm{c}}^2}x]$

a
(a) $y = {(x{\cot ^3}x)^{3/2}}$

$\therefore \frac{{dy}}{{dx}} = \frac{3}{2}{(x{\cot ^3}x)^{1/2}}[{\cot ^3}x + 3x{\cot ^2}x( - {\rm{cose}}{{\rm{c}}^2}x)]$

$ = \frac{3}{2}{(x{\cot ^3}x)^{1/2}}[{\cot ^3}x - 3x{\cot ^2}x\,{\rm{cose}}{{\rm{c}}^2}x]$.

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