If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$, where $\mathrm{a}>\mathrm{b}>0$, then $\frac{\mathrm{d} x}{\mathrm{~d} y}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is

If $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$, then $\frac{\mathrm{d}^2 y}{d x^2}$ is equal to

If $\mathrm{f}(x)=\log _{x^2}(\log x)$, then at $x=\mathrm{e}, \mathrm{f}^{\prime}(x)$ has the value

Let $\mathrm{f}(x)=\frac{x}{\sqrt{\mathrm{a}^2+x^2}}-\frac{\mathrm{d}-x}{\sqrt{\mathrm{~b}^2+(\mathrm{d}-x)^2}}, x \in \mathbb{R}$ where $\mathrm{a}, \mathrm{b}, \mathrm{d}$ are non-zero real constants. Then