1. ## Derivative

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

$\displaystyle 0 \leq y < \pi$

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

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

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

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

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

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

$\displaystyle 2x = secy$

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

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

Can I rearrange and replace it like this?

So I'll get, ;

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

cancel the two's and replace sec^2y

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

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

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!

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

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

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

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

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

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

$\displaystyle 2x = secy$

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

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

Can I rearrange and replace it like this?

So I'll get, ;

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

cancel the two's and replace sec^2y

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

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

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:

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

and therefore

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

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

which simplifies to the given result.

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

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

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

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

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

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

$\displaystyle 2x = secy$

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

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

Can I rearrange and replace it like this?

So I'll get, ;

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

cancel the two's and replace sec^2y

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

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

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

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

Originally Posted by Tweety
$\displaystyle \color{red} 2x = secy$
This is wrong.
You probably multiply this equation $\displaystyle x=sec(\frac{y}{2})$ 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.

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

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

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

Thank you!

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

is not true since
you have this

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

not this

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

first step is true

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

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

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

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

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

sub sec y/2 and tan y/2

$\displaystyle \frac{dy}{dx} = \frac{2}{x\sqrt{x^2-1}}$

5. Thanks alot, I understand where I went wrong.

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

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

Just to clarify, writting $\displaystyle sec\frac{y}{2}$ is not the same as writting $\displaystyle \frac{secy}{2}$ ?
In general no. To write unambiguously use brackets. Compare my answer.