Question

differentiate w.r.t.x.
y=cosec^-1[10/6sin(2^x)-8cos(2^x)]​

Answer 1

we have to differentiate $y=cosec^{-1}\left[\frac{10}{6sin(2^x)-8cos(2^x)}\right]$ with respect to x.

solution : $y=cosec^{-1}\left[\frac{10}{6sin(2^x)-8cos(2^x)}\right]$

differentiating y with respect to x.

y' = $\frac{d\left{cosec^{-1}\left[\frac{10}{6sin(2^x)-8cos(2^x)}\right]\right}}{dx}$

let $y=cosec^{-1}\left[\frac{10}{6sin(2^x)-8cos(2^x)}\right]$ = P

y = $\frac{1}{|p|\sqrt{p^2-1}}\frac{dp}{dx}$

= $\frac{6sin(2^x)-8cos(2^x)}{10}\frac{1}{\sqrt{\left(\frac{10}{6sin(2^x)-8cos(2^x)}\right)^2-1}}\frac{d\left(\frac{10}{6sin(2^x)-8cos(2^x)}\right)}{dx}$

= $\frac{(6sin(2^x)-8cos(2^x))^2}{10}\frac{1}{\sqrt{100-(6sin(2^x)-8cos(2^x))^2}}\times10(6sin(2^x)-8cos(2^x))^{-2}\frac{d(6sin(2^x)-8cos(2^x))}{dx}$

= $\frac{6cos(2^x)2^xlog2+8sin(2^x)2^xlog2}{\sqrt{100-(6sin(2^x)-8cos(2^x))^2}}$

= $\frac{2^xlog2(6cos(2^x)+8sin(2^x))}{\sqrt{100-(6sin(2^x)-8cos(2^x))^2}}$