If $x, y, z$ are in Arithmetic Progression and $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are also in Arithmetic progression, where $x, z>0$ and $x z<1, y<1$, then

Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ w.r.t. $\cos ^{-1}\left(4 x^3-3 x\right)$ is

If $\frac{\mathrm{d}}{\mathrm{d} x} \mathrm{f}(x)=4 x^3-\frac{3}{x^4}$ such that $\mathrm{f}(2)=0$, then $\mathrm{f}(x)$ is equal to

$$\lim _\limits{x \rightarrow 0} \frac{9^x-4^x}{x\left(9^x+4^x\right)}=$$