Question

Let $f\left(x\right)=g\left(x\right)cos2x+x^2-5,$ where $g\left(x\right)$ is a twice differentiable function or $R$ such that $g^{\prime }\left(\frac{\pi }{4}\right)=-1,$ then $f"\left(\frac{\pi }{4}\right)$ is equal to

Answer 1

Differentiate it w.r.t. $x$
$f^{\prime }\left(x\right)=g^{\prime }\left(x\right).\cos 2x-2\sin 2x.g\left(x\right)+2x$
$f"\left(x\right)=g"\left(x\right)\cos 2x-2\sin 2x.g^{\prime }\left(x\right)-4\cos 2x.g\left(x\right)-2\sin 2x.g^{\prime }\left(x\right)+2$
put $x=\pi /4\Rightarrow f^{\prime \prime }\left(\frac{\pi }{4}\right)=0-2g^{\prime }\left(\pi /4\right)-2g^{\prime }\left(\pi /4\right)+2=6$