Question

If y= sin inverse(3sinx+4cosx)/5 then find dy/dx

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=sin^{-1}\bigg(\dfrac{3\,sin(x)+4\,cos(x)}{5}\bigg)}$

$\sf{\implies\,y=sin^{-1}\bigg(\dfrac{3}{5}\,sin(x)+\dfrac{4}{5}\,cos(x)\bigg)}$

$\sf{Let\,\,tan(\alpha)=\dfrac{4}{3}}\\\sf{\implies\,sin(\alpha)=\dfrac{4}{5}\,\,\,\,and\,\,\,\,\,cos(\alpha)=\dfrac{3}{5}}$

So,

$\sf{\,y=sin^{-1}\{sin(x)cos(\alpha)+cos(x)sin(\alpha)\}$

$\sf{\implies\,y=sin^{-1}\{sin(x+\alpha)\}$

$\sf{\implies\,y=x+\alpha$

Differentiating both sides w.r.t x,

$\sf{\implies\,\dfrac{dy}{dx}=1$