# Math Help - problem with limit & derivative

1. ## 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 :

img.top {vertical-align:15%;}

$X^2+3xy+2y^2 = 9$
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

2. ## Re: problem with limit & derivative

Originally Posted by mido22
$\lim_{h\to0}\frac{\cos x-\cos(x+h)}h$
Use the sum and difference formula for cosine.

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

Originally Posted by mido22
find dy/dx :
$x^2+3xy+2y^2 = 9$
Differentiate implicitly and then solve for $\frac{dy}{dx}.$ If you're really stuck, show us what you've tried.

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 $y=\tan\frac x2$ whose tangent lines are perpendicular to the given line, or what?

3. ## Re: problem with limit & derivative

for the third problem ..yes i 'm looking for points on $y=tan x/2$ whose tangent perpen. to 3x=8-3y
here is my attempt for the 2nd one :
$y^2 = 9 - 3xy - x^2$
$y = +- \sqrt{9-3xy-x^2}$
$dy/dx = +-\frac{2x+3y+(3x*(dy/dx))}{2*\sqrt{9-3xy-x^2}}$

4. ## Re: problem with limit & derivative

Originally Posted by mido22
$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 $y$ first. That isn't really necessary:

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

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

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

And continue...

5. ## Re: problem with limit & derivative

Originally Posted by mido22
for the third problem ..yes i 'm looking for points on $y=tan x/2$ whose tangent perpen. to 3x=8-3y
What do we know about the slopes of perpendicular lines?

6. ## 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
$\frac{dy}{dx} = 0.5 * \sec^2\frac{x}{2} = 1$
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?

7. ## Re: problem with limit & derivative

and i get the solution of the 2nd problem
$2x+3y+3x\frac{dy}{dx}+4y\frac{dy}{dx}$

8. ## Re: problem with limit & derivative

Hello, mido22!

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

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

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

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

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

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

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

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