d dx ( a^ _ 10 cose c ^ - 1 x )= - Vidyadip

MCQ

$\frac{d}{{dx}}\left( {{a^{{{\log }_{10}}{\rm{cose}}{{\rm{c}}^{ - 1}}x}}} \right)$=

A
${a^{{{\log }_{10}}\cos e{c^{ - 1}}x}}.\frac{1}{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{x\sqrt {{x^2} - 1} }}.{\log _{10}}a$
✓
$ - {a^{{{\log }_{10}}\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{|x|\sqrt {{x^2} - 1} }}.{\log _{10}}a$
C
${a^{{{\log }_{10}}\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{|x|\sqrt {{x^2} - 1} }}.{\log _{10}}a$
D
$ - {a^{{{\log }_{10}}\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{x\sqrt {{x^2} - 1} }}.{\log _{10}}a$

✓

Answer

Correct option: B.

$ - {a^{{{\log }_{10}}\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{|x|\sqrt {{x^2} - 1} }}.{\log _{10}}a$

b
(b) Let ${\rm{cose}}{{\rm{c}}^{ - 1}}x = u,\,\,{\log _{10}}u = v$
By chain rule, $\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}}.\frac{{dv}}{{du}}.\frac{{du}}{{dx}}$
$ = {a^v}.{\log _e}a \times \frac{1}{u}.{\log _{10}}e \times \frac{1}{{ - |x|\sqrt {{x^2} - 1} }}$
$ = - {a^{{{\log }_{10}}co{\rm{se}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{{\rm{cose}}{{\rm{c}}^{ - 1}}x}}.\frac{1}{{|x|\sqrt {{x^2} - 1} }}.{\log _{10}}a$.

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