# differention

• Jul 6th 2013, 07:32 PM
srirahulan
differention
If ,y=$\displaystyle \ln{secx}, |x|< \frac{\pi}{2}$ show that $\displaystyle \frac{d^2y}{dx^2}=e^{2y}$
• Jul 6th 2013, 07:45 PM
Prove It
Re: differention
It doesn't...

\displaystyle \displaystyle \begin{align*} y &= \ln{ \left[ \sec{(x)} \right] } \\ &= \ln{ \left\{ \left[ \cos{(x)} \right] ^{-1} \right\} } \\ &= -\ln{ \left[ \cos{(x)} \right] } \end{align*}

Now let \displaystyle \displaystyle \begin{align*} u = \cos{(x)} \implies y = -\ln{(u)} \end{align*}, then

\displaystyle \displaystyle \begin{align*} \frac{du}{dx} &= -\sin{(x)} \\ \\ \frac{dy}{du} &= -\frac{1}{u} \\ &= -\frac{1}{\cos{(x)}} \\ \\ \frac{dy}{dx} &= -\sin{(x)} \left[ -\frac{1}{\cos{(x)}} \right] \\ &= \frac{\sin{(x)}}{\cos{(x)}} \\ \\ \frac{d^2y}{dx^2} &= \frac{\cos{(x)}\cos{(x)} - \sin{(x)} \left[ -\sin{(x)} \right] }{\left[ \cos{(x)} \right] ^2} \\ &= \frac{\cos^2{(x)} + \sin^2{(x)}}{\cos^2{(x)}} \\ &= \frac{1}{\cos^2{(x)}} \\ &= \sec^2{(x)} \end{align*}

Now remembering that \displaystyle \displaystyle \begin{align*} y = \ln{ \left[ \sec{(x)} \right] } \end{align*} that means \displaystyle \displaystyle \begin{align*} \sec{(x)} = e^y \end{align*}, and so \displaystyle \displaystyle \begin{align*} \frac{d^2y}{dx^2} = e^y \end{align*}, NOT \displaystyle \displaystyle \begin{align*} e^{2y} \end{align*}.
• Jul 6th 2013, 08:23 PM
Soroban
Re: differention
Hello, srirahulan!

Quote:

$\displaystyle \text{If }\,y\:=\:\ln(\sec x),\;|x|< \tfrac{\pi}{2},\;\text{ show that: }\:\frac{d^2y}{dx^2}\:=\:e^{2y}$

We have: .$\displaystyle y \:=\:\ln(\sec x)\;\;[1]$

Then: .$\displaystyle \frac{dy}{dx} \:=\:\frac{1}{\sec x}(\sec x\tan x) \:=\:\tan x$

Hence: .$\displaystyle \frac{d^2y}{dx^2} \:=\:\sec^2\!x\;\;[2]$

From [1], we have: .$\displaystyle y \:=\:\ln(\sec x) \quad\Rightarrow\quad \sec x \:=\:e^y$

Substitute into [2]: .$\displaystyle \frac{d^2y}{dx^2} \:=\:(e^y)^2$

. . . . . . Therefore: .$\displaystyle \frac{d^2y}{dx^2} \:=\:e^{2y}$
• Jul 7th 2013, 12:53 AM
Prove It
Re: differention
It appears I am unable to edit my previous post. I for some reason forgot that the second derivative is \displaystyle \displaystyle \begin{align*} \sec^2{(x)} \end{align*}, not \displaystyle \displaystyle \begin{align*} \sec{(x)} \end{align*}.
• Jul 7th 2013, 03:50 AM
Isomorphism
Re: differention
Another way is:

We can write $\displaystyle y = \ln(\sec x)$ as $\displaystyle e^y = \sec x$.

Now differentiate both sides:

$\displaystyle e^y \frac{dy}{dx}= \sec x \tan x = e^y \tan x$

Cancelling $\displaystyle e^y$ on both sides, we get $\displaystyle \frac{dy}{dx} = \tan x$.

Differentiating this equation we get,

$\displaystyle \frac{d^2y}{dx^2} = \sec^2 x = (e^y)^2 = e^{2y}$.