If $x = y\sqrt{1 - y^2}$,then $\frac{dy}{dx} = $

Let $f: R \rightarrow R$ be a differentiable function and $f(1)=4$. Then the value of $\lim _{x \rightarrow 1} \int_4^{f(x)} \frac{2 t}{x-1} dt$, if $f^{\prime}(1)=2$ is

If $\log _{10}\left(\frac{x^{3}-y^{3}}{x^{3}+y^{3}}\right)=2$,then $\frac{dx}{dy} = $

If $y = (1 + \frac{1}{x}) (1 + \frac{2}{x}) (1 + \frac{3}{x}) . . . . . . (1 + \frac{n}{x})$ and $x \neq 0$. When $x = -1$,find $\frac{dy}{dx}$.

If $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{dy}{dx}=$

Let $f$ be a differentiable function such that $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$ and $f(1) = 2$. Then the area enclosed by $\frac{|f(x) - x|^{1/3}}{17} + \frac{|f(y) - y|^{1/3}}{2} \le \frac{1}{4}$ is