# Derivative

• January 17th 2010, 04:31 AM
Tweety
Derivative
Quote:

Given that $x = sec\frac{y}{2}$

$0 \leq y < \pi$

show that $\frac{dy}{dx} = \frac{2}{x \sqrt{x^{2}-1}}$
$\frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}$

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

$sec y = 2x$ so replace $\frac{secy}{2}$ with 2x

and using the indentity $1 + tan^{2}y = sec^{2}y$

$sec^{2}y - 1 = tan^{2}y$

$(\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4}$

$2x = secy$

$(2x)^{2} = sec^{2}y$

$4x^{2} = sec^{2}y$

Can I rearrange and replace it like this?

So I'll get, ;

$\frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}}$

cancel the two's and replace sec^2y

$\frac{1}{\frac{1}{2} \times x \times \frac{4x^{2}-1}{4}}$

so after simplifying I get, $\frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}}$

Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

Thank you!
• January 17th 2010, 05:01 AM
earboth
Quote:

Originally Posted by Tweety
$\frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}$

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

$sec y = 2x$ so replace $\frac{secy}{2}$ with 2x

and using the indentity $1 + tan^{2}y = sec^{2}y$

$sec^{2}y - 1 = tan^{2}y$

$(\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4}$

$2x = secy$

$(2x)^{2} = sec^{2}y$

$4x^{2} = sec^{2}y$

Can I rearrange and replace it like this?

So I'll get, ;

$\frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}}$

cancel the two's and replace sec^2y

$\frac{1}{\frac{1}{2} \times x \times \frac{4x^{2}-1}{4}}$

so after simplifying I get, $\frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}}$

Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

Thank you!

My attempt:

$x=\sec\left(\frac12 y\right)~\implies~\cos\left(\frac12 y\right)=\frac1x$

and therefore

$y = 2\arccos\left(\frac1x\right)$

$\frac{dy}{dx}=2\left(-\dfrac{1}{\sqrt{1-\left(\frac1x \right)^2}} \right) \cdot \left(-\frac1{x^2}\right) = 2\left(\dfrac{1}{\frac1x \cdot \sqrt{x^2-1}} \right) \cdot \left(\frac1{x^2}\right)$

which simplifies to the given result.
• January 17th 2010, 05:02 AM
General
Quote:

Originally Posted by Tweety
$\frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}$

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

$sec y = 2x$ so replace $\frac{secy}{2}$ with 2x

and using the indentity $1 + tan^{2}y = sec^{2}y$

$sec^{2}y - 1 = tan^{2}y$

$(\frac{tany}{2})^{2} = \frac{sec^{2}y -1}{4}$

$2x = secy$

$(2x)^{2} = sec^{2}y$

$4x^{2} = sec^{2}y$

Can I rearrange and replace it like this?

So I'll get, ;

$\frac{1}{ \frac{1}{2}\times \frac{2x}{2} \times \frac{sec^{2}y-1}{4}}$

cancel the two's and replace sec^2y

$\frac{1}{\frac{1}{2} \times x \times \frac{4x^{2}-1}{4}}$

so after simplifying I get, $\frac{1}{\frac{x}{2} \times \frac{4x^{2}-1}{4}}$

Can someone please help me from here, as I have tired to rearrange it but don't get the desired expression.

And I am not even sure what I have done so far is correct, which is probably why I am not getting the right expression?

Thank you!

The identity is:
$sec^2(\theta) - 1 = tan^2(\theta)$
For any $\theta$.
substitute $\theta=\frac{y}{2}$
$sec^2(\frac{y}{2}) - 1 = tan^2(\frac{y}{2})$
since $sec(\frac{y}{2}) = x$, then the last one can be written as:
$x^2 - 1 = tan^2(\frac{y}{2})$
Hence:
$tan(\frac{y}{2})=\sqrt{x^2-1}$
Not substitute this in $\frac{dy}{dx} = \frac{1}{ \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}}$

------------

Quote:

Originally Posted by Tweety
$\color{red} 2x = secy$

This is wrong.
You probably multiply this equation $x=sec(\frac{y}{2})$ by 2, and you canceled the 2s from the right side.
Which is wrong, one of this 2s is inside the angle of a trigonometric function.
You can not cancel it with another 2 which outside the angle.
• January 17th 2010, 05:03 AM
Amer
Quote:

Originally Posted by Tweety
$\frac{dx}{dy} = \frac{1}{2} sec\frac{y}{2} tan\frac{y}{2}$

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

$sec y = 2x$ so replace $\frac{secy}{2}$ with 2x

Thank you!

this $sec y = 2x$ so replace $\frac{secy}{2}$ with 2x

is not true since
you have this

$\sec \frac{y}{2} = x$

not this

$\frac{\sec y}{2} = x$

first step is true

$\frac{dy}{dx} = \frac{2}{\sec \frac{y}{2} \tan \frac{y}{2}}$

want to find tan y/2 with respect to x
you have

$\sec \frac{y}{2} = x$

$\sec ^2 \frac{y}{2} = x^2$

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

sub sec y/2 and tan y/2

$\frac{dy}{dx} = \frac{2}{x\sqrt{x^2-1}}$
• January 17th 2010, 05:45 AM
Tweety
Thanks alot, I understand where I went wrong.

Just to clarify, writting $sec\frac{y}{2}$ is not the same as writting $\frac{secy}{2}$ ?
• January 17th 2010, 05:48 AM
earboth
Quote:

Originally Posted by Tweety
Thanks alot, I understand where I went wrong.

Just to clarify, writting $sec\frac{y}{2}$ is not the same as writting $\frac{secy}{2}$ ?

In general no. To write unambiguously use brackets. Compare my answer.