# problem with limit & derivative

• Jun 11th 2012, 09:39 AM
mido22
problem with limit & derivative
i have problems with limit and derivative
first problem :
Code:

l i m      Cos X - Cos(X+h) h-->0  ---------------------                   h
second one
Code:

find dy/dx : $\displaystyle X^2+3xy+2y^2 = 9$
third one :
Code:

find points on curve y=tan(x/2) , x belong to ]0,360[ and the tangent is perpendicular to straight line 3x=8-3y i got 4 solutions in this problem .... 2 of them i think refused
• Jun 11th 2012, 10:04 AM
Reckoner
Re: problem with limit & derivative
Quote:

Originally Posted by mido22
$\displaystyle \lim_{h\to0}\frac{\cos x-\cos(x+h)}h$

Use the sum and difference formula for cosine.

Alternatively, recognize that $\displaystyle \lim_{h\to0}\frac{\cos x-\cos(x+h)}h=-\lim_{h\to0}\frac{\cos(x+h)-\cos x}h=-\frac d{dx}\left[\cos x\right].$

Quote:

Originally Posted by mido22
find dy/dx :
$\displaystyle x^2+3xy+2y^2 = 9$

Differentiate implicitly and then solve for $\displaystyle \frac{dy}{dx}.$ If you're really stuck, show us what you've tried.

Quote:

Originally Posted by mido22
find points on curve y=tan(x/2) , x belong to ]0,360[ and the tangent is perpendicular to straight line 3x=8-3y
i got 4 solutions in this problem .... 2 of them i think refused

I'm not even sure what this is saying. Are you looking for points on $\displaystyle y=\tan\frac x2$ whose tangent lines are perpendicular to the given line, or what?
• Jun 11th 2012, 10:17 AM
mido22
Re: problem with limit & derivative
for the third problem ..yes i 'm looking for points on $\displaystyle y=tan x/2$ whose tangent perpen. to 3x=8-3y
here is my attempt for the 2nd one :
$\displaystyle y^2 = 9 - 3xy - x^2$
$\displaystyle y = +- \sqrt{9-3xy-x^2}$
$\displaystyle dy/dx = +-\frac{2x+3y+(3x*(dy/dx))}{2*\sqrt{9-3xy-x^2}}$
• Jun 11th 2012, 10:25 AM
Reckoner
Re: problem with limit & derivative
Quote:

Originally Posted by mido22
$\displaystyle y^2 = 9 - 3xy - x^2 y = +- sqrt(9-3xy-x^2) dy/dx = +- (2x+3y+(3x*(dy/dx)))/(2*sqrt(9-3xy-x^2))$

That's a mess, but it looks like you're trying to solve for $\displaystyle y$ first. That isn't really necessary:

$\displaystyle x^2+3xy+2y^2 = 9$

$\displaystyle \Rightarrow\frac d{dx}\left[x^2+3xy+2y^2\right] = \frac d{dx}[9]$

$\displaystyle \Rightarrow\frac d{dx}\left[x^2\right]+3\frac d{dx}\left[xy\right]+2\frac d{dx}\left[y^2\right] = 0$

And continue...
• Jun 11th 2012, 10:26 AM
Reckoner
Re: problem with limit & derivative
Quote:

Originally Posted by mido22
for the third problem ..yes i 'm looking for points on $\displaystyle y=tan x/2$ whose tangent perpen. to 3x=8-3y

What do we know about the slopes of perpendicular lines?
• Jun 11th 2012, 10:34 AM
mido22
Re: problem with limit & derivative
the perpen . line equation : 3x=8-3y so its slope is -1
so the slope of tangent is -1 / -1 = 1
$\displaystyle \frac{dy}{dx} = 0.5 * \sec^2\frac{x}{2} = 1$
then i get x/2 =90 or 315 or 135 or 450
then i cancelled the 2nd and fourth solutions becz x belong to ]0,360[

is this right?
• Jun 11th 2012, 10:39 AM
mido22
Re: problem with limit & derivative
and i get the solution of the 2nd problem
$\displaystyle 2x+3y+3x\frac{dy}{dx}+4y\frac{dy}{dx}$
• Jun 11th 2012, 02:40 PM
Soroban
Re: problem with limit & derivative
Hello, mido22!

Quote:

Find points on curve $\displaystyle y\,=\,\tan\left(\tfrac{x}{2}\right),\; x \in [0^o,\,360^o]$
and the tangent is perpendicular to straight line $\displaystyle 3x\,=\,8-3y$

The straight line is: .$\displaystyle y \:=\:-x + \tfrac{8}{3}$
. . Its slope is $\displaystyle -1.$

The slope of the tangent is: .$\displaystyle y' \:=\:\tfrac{1}{2}\sec^2(\tfrac{x}{2})$

We want the slope of the tangent to be $\displaystyle +1.$

. . $\displaystyle \tfrac{1}{2}\sec^2(\tfrac{x}{2}) \:=\:1 \quad\Rightarrow\quad \sec^2(\tfrac{x}{2}) \:=\:2 \quad\Rightarrow\quad \sec(\tfrac{x}{2}) \:=\:\pm\sqrt{2}$

. . $\displaystyle \tfrac{x}{2} \:=\:45^o,\:135^o,\:225^o,\:315^o\:\hdots$

. . $\displaystyle x \:=\:90^o,\:270^o,\:450^o,\:630^o\:\hdots$

Therefore: .$\displaystyle x \;=\;90^o,\:270^o$