If $y=(x+1)(x^2+1)(x^4+1)(x^8+1)$,then $\lim _{x \rightarrow-1} \frac{dy}{dx}=$

The equation of a tangent to the curve $y \cot x = y^3 \tan x$ at the point where the abscissa is $\frac{\pi}{4}$ is:

The function defined by $f(x) = \begin{cases} |x - 3|, & x \ge 1 \\ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{13}{4}, & x < 1 \end{cases}$ is

Two curves $C_1 : y = x^2 - 3$ and $C_2 : y = kx^2, k \in R$,intersect each other at two different points. The tangent drawn to $C_2$ at one of the points of intersection $A \equiv (a, y_1), (a > 0)$ meets $C_1$ again at $B(1, y_2), (y_1 \neq y_2)$. The value of '$a$' is

Observe the following statements:
$I. f(x) = a x^{41} + b x^{-40} \Rightarrow \frac{f^{\prime \prime}(x)}{f(x)} = 1640 x^{-2}$
$II. \frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^2}\right) = \frac{1}{1+x^2}$
Which of the following is correct?

$f(x)$ is a differentiable function and given $f^{\prime}(2)=6$ and $f^{\prime}(1)=4$, then $L=\lim _{h \rightarrow 0} \frac{f\left(2+2 h+h^2\right)-f(2)}{f\left(1+h-h^2\right)-f(1)}$