# How do you find (d^2y)/(dx^2) given x=tany?

Sep 5, 2016

$\forall y \in \mathbb{R} , \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$.

#### Explanation:

$x = \tan y$.

Case : 1 $\text{Suppose that,} y \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) \Rightarrow y = a r c \tan x , x \in \mathbb{R}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$.

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{1 + {x}^{2}}\right)$

Since, $\frac{d}{\mathrm{dt}} \left(\frac{1}{t}\right) = - \frac{1}{t} ^ 2$, we have, by the Chain Rule,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{{\left(1 + {x}^{2}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right) = \frac{- 2 x}{{\left(1 + {x}^{2}\right)}^{2}}$

Case : 2 $\text{Next, let } y \in \left(\frac{\pi}{2} , 3 \frac{\pi}{2}\right) \Rightarrow \frac{\pi}{2} < , y < , 3 \frac{\pi}{2}$

$\Rightarrow \frac{\pi}{2} - \pi < , y - \pi < , 3 \frac{\pi}{2} - \pi$

rArr -pi/2<,y-pi<,pi/2, &, tan(y-pi)=-tan(pi-y)=-(-tany)=tany=x.

Thus, in this Case, $x = \tan \left(y - \pi\right) , w h e r e , \left(y - \pi\right) \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$\therefore \text{ by defn. of arc tan, } y - \pi = a r c \tan x , \mathmr{and} , y = \pi + a r c \tan x$

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$, as in Case : 1

Thus, $\forall y \in \mathbb{R} , \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$.

Sep 5, 2016

To differentiate without using the inverse tangent, see below.

#### Explanation:

$\tan y = x$

Differentiate both sides with respect to $x$.

$\frac{d}{\mathrm{dx}} \left(\tan x\right) = \frac{d}{\mathrm{dx}} \left(x\right)$

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} y$

Differentiate again w.r.t. $x$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 \cos y \left(- \sin y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Now rfeplace $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2 \sin y {\cos}^{3} y$

To see that this is the same as the other answer

$\tan y = x$ $\Rightarrow$ $\cos y = \frac{1}{1 + {x}^{2}}$

(To see this draw and label a right triangle with angle $y$, opposite side $x$ and adjacent side $1$. So the hypotenuse is $1 + {x}^{2}$.)

(Or use $\tan y = x$ $\Rightarrow$ ${\tan}^{2} y = {x}^{2}$, so $1 + {\tan}^{2} y = 1 + {x}^{2}$ and ${\sec}^{y} = 1 + {x}^{2}$ so that $\cos y = \frac{1}{1 + {x}^{2}}$

Now,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2 \sin y {\cos}^{3} y$

$= - 2 \sin \frac{y}{\cos} y {\cos}^{4} y$

$= - 2 \tan y {\left({\cos}^{2} y\right)}^{2}$

$= - 2 x \frac{1}{1 + {x}^{2}} ^ 2$

$= \frac{- 2 x}{1 + {x}^{2}} ^ 2$.