Question

$\frac{d}{dx}${${\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$}=

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{-1}{\sqrt{1+x^2}}$
• C. $\frac{2}{\sqrt{1-x^2}}$
• D. $0$

Answer 1

The correct option is A $\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}\int {\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$

Let $x=\sin \theta$

$dx=cos\theta d\theta$

$\therefore \frac{d}{dx}\left({\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)\right)$

$\frac{d}{dx}\left(si{n}^{-1}\left(\frac{5sin\theta +12cos\theta }{13}\right)\right)$

$=sec\theta \frac{d}{d\theta }\left({\sin }^{-1}\left(\frac{5sin\theta +12cos\theta }{13}\right)\right)$

$=sec\theta \left(\frac{1}{\sqrt{1-{\left(\frac{5sin\theta +12cos\theta }{13}\right)}^2}}\right)\frac{d}{d\theta }\left(\frac{5sin\theta +12cos\theta }{13}\right)$

$=sec\theta \left(\frac{13}{\sqrt{169-(5sin\theta +12cos\theta {)}^2}}\right)\times (5cos\theta -12sin\theta )$

$=\frac{sec\theta }{\sqrt{169-(5sin\theta +12cos\theta {)}^2}}\times (5cos\theta -12sin\theta )$

$=\frac{sec\theta \times (5cos\theta -12sin\theta )}{\sqrt{169-(25si{n}^2\theta +144co{s}^2\theta +120sin\theta cos\theta }}$

$=\frac{sec\theta \times \left(5cos\theta \right)-sec\theta \times 12sin\theta }{\sqrt{144-119co{s}^2\theta -120sin\theta cos\theta }}$

$=\frac{5-12tan\theta }{\sqrt{144-119co{s}^2\theta -120sin\theta cos\theta }}$

$=\frac{1}{\sqrt{1-x^2}}$