Question

let f(x)=sinx , g(x)=[x+1] and g(f(x))=h(x) then $h^{{}^{\prime }}\left(\frac{\pi }{2}\right)$

Answer 1

$f\left(x\right)=\sin xg\left(x\right)=[x+1]h\left(x\right)=g\left(f\right(x\left)\right)=[\sin x+1]=\left[\sin x\right]+1h^{{}^{\prime }}\left(x\right)=\left[\cos x\right]+0h^{{}^{\prime }}\left(\frac{\pi }{2}\right)=\left[\cos \frac{\pi }{2}\right]=0$