If $f(x) = \sec (3x)$, then $f'\left (\dfrac {3\pi}{4}\right ) =$

If the functions $\displaystyle f\left ( x \right )=\sin \left ( x+a \right )$ and $\displaystyle g\left ( x \right )=b\sin x+c\cos x$ satisfy $\displaystyle f\left ( 0 \right )=g\left ( 0 \right )$ and $\displaystyle {f}'\left ( 0 \right )={g}'\left ( 0 \right )$ then

$\displaystyle\frac{dy}{dx}$ at $\displaystyle t=\frac{\pi}{4}$ for $\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$ and $y=a\sin{t}$ is

If the prime sign (') represents differentiation w.r.t. $x$ and $f^{'}=\sin x+\sin 4x.\cos x$, then $f^{'}\left ( 2x^{2}+\cfrac{\pi }{2} \right )$ at $x=\sqrt{\dfrac{\pi }{2}}$ is equal to

y=log(secx) হলে dy/dx=কত?