Question

The arrangement of the following curves in the ascending order of slopes of their tangents at the given points.
$A)y=\frac{1}{1+x^2}$ at $x=0$
$B)y=2e^{\frac{-x}{4}},$ where it cuts the y-axis
$C)y=cos(x)$ at $x=\frac{-\pi }{4}$
$D)y=4x^2$ at $x=-1$

• A. DCBA
• B. ACBD
• C. ABCD
• D. DBAC

Answer 1

The correct option is D DBAC

(A) $y^1=-\frac{2x}{(1+x^2)}$
$y^1{|}_{x=0}=0$
(B) $y^1=-\frac{2x}{(4}$
$y^1{|}_{x=0}=-\frac{1}{2}$
(C) $y^1=-sinx$
$y^1{|}_{x{=}^{-pi}{/}_4}=\frac{1}{\sqrt{2}}$
(D) $y^1=8x$
$y^1{|}_{x=-1}=-8$
$\left(C\right)>\left(A\right)>\left(B\right)>\left(D\right)$