Question

If $f\left(x\right)=\sin 2x-\cos 2x,$ find $f^{\prime }\left(\frac{\pi }{6}\right)$

Answer 1

We have,

$f\left(x\right)=\sin 2x-\cos 2x$

On differentiating both sides w.r.t x, we get

$f^{\prime }\left(x\right)=\cos 2x\times \left(2\right)-(-\sin 2x)\times \left(2\right)$

$f^{\prime }\left(x\right)=2(\cos 2x+\sin 2x)$

Put $x=\frac{\pi }{6}$

Therefore,

$f^{\prime }\left(\frac{\pi }{6}\right)=2(\cos \frac{\pi }{3}+\sin \frac{\pi }{3})$

$f^{\prime }\left(\frac{\pi }{6}\right)=2(\frac{1}{2}+\frac{\sqrt{3}}{2})$

$f^{\prime }\left(\frac{\pi }{6}\right)=2\left(\frac{1+\sqrt{3}}{2}\right)$

$f^{\prime }\left(\frac{\pi }{6}\right)=1+\sqrt{3}$

Hence, the value is $1+\sqrt{3}$.