differentiation help..

**omibayne** · October 25th 2008, 09:26 PM

The question is:

Find slopes of all lines tangent to the graph of the relation:

y^3 + 15x^15y^2 - 50y = 150 - 150x^5

at the points on its graph where x=1

**Jhevon** · October 25th 2008, 09:37 PM

Originally Posted by omibayne

The question is:

Find slopes of all lines tangent to the graph of the relation:

y^3 + 15x^15y^2 - 50y = 150 - 150x^5

at the points on its graph where x=1

there are three points for x = 1: $(1,0),~ \left( 1, \frac {-15 + 5 \sqrt{17}}2\right),~\text{and } \left( 1, \frac {-15 - 5 \sqrt{17}}2 \right)$

differentiate implicitly with respect to x and plug each of these points in to find the slopes of each of the three tangent lines.

**Jhevon** · October 25th 2008, 09:45 PM

Originally Posted by omibayne

the correct answers for slopes are :- 15, -35 and -55 but i dont know how to get them :S:S

i just told you how to get them

you do know how to differentiate implicitly, right?

**omibayne** · October 25th 2008, 09:48 PM

I do know but it didnt work...:S

**Jhevon** · October 25th 2008, 09:56 PM

Originally Posted by omibayne

I do know but it didnt work...:S

$y^3 + 15x^{15}y^2 - 50y = 150 - 150x^5$

$\Rightarrow \frac {dy}{dx} = \frac {-(750x^4 + 225x^{14}y^2)}{3y^2 + 30x^{15}y - 50}$

now plug in the points given earlier

**omibayne** · October 25th 2008, 10:29 PM

can smn help me differentiate this equation. i tried it several times but it didnt work..any help will be appreciated..thanks

y^3 + 5x^15y^2 - 50y = 150 - 150x^5

**11rdc11** · October 25th 2008, 10:37 PM

Originally Posted by omibayne

can smn help me differentiate this equation. i tried it several times but it didnt work..any help will be appreciated..thanks

y^3 + 5x^15y^2 - 50y = 150 - 150x^5

Can you show us what work you have so far?

$\frac{dy}{dx}=\frac{-750x^4-75x^{14}y^2}{3y^2+10x^{15}y-50}$

**11rdc11** · October 25th 2008, 10:50 PM

Originally Posted by Jhevon

double posting is against the rules. and it is disrespectful to do so when someone already tried to help you

I agree with Jhevon but did you repost because you posted the problem wrong? If so just edit your original post and don't start a new thread.

One post has

y^3 + 15x^15y^2 - 50y = 150 - 150x^5

and the other has

y^3 + 5x^15y^2 - 50y = 150 - 150x^5

**omibayne** · October 25th 2008, 10:52 PM

[/tex]

Originally Posted by 11rdc11

Can you show us what work you have so far?

$\frac{dy}{dx}=\frac{3y^2+5x^{15}2y-50}{-750x^4-75x^{14}y^2}$

i got $\frac{dy}{dx}=\frac{-750x^4-75x^{14}y^2}{3y^2+10x^{15}y-50}$

**omibayne** · October 25th 2008, 10:55 PM

Originally Posted by Jhevon

double posting is against the rules. and it is disrespectful to do so when someone already tried to help you

Originally Posted by 11rdc11

I agree with Jhevon but did you repost because you posted the problem wrong? If so just edit your original post and don't start a new thread.

One post has

y^3 + 15x^15y^2 - 50y = 150 - 150x^5

and the other has

y^3 + 5x^15y^2 - 50y = 150 - 150x^5

yeah i posted the wrong question...i am new to this so sorry abt that...

**11rdc11** · October 25th 2008, 10:55 PM

Originally Posted by omibayne

[/tex]

i got $\frac{dy}{dx}=\frac{-750x^4-75x^{14}y^2}{3y^2+10x^{15}y-50}$

Your correct I made a mistake when typing it into latex. I edited my post

**11rdc11** · October 25th 2008, 10:58 PM

Originally Posted by omibayne

yeah i posted the wrong question...i am new to this so sorry abt that...

It's kewl. We were all newbies at one point lol. Next time just make the correction to your original post ok

**omibayne** · October 25th 2008, 11:01 PM

Originally Posted by 11rdc11

It's kewl. We were all newbies at one point lol. Next time just make the correction to your original post ok

ok thanx

**toraj58** · October 25th 2008, 11:15 PM

as a genral solution you can sovle all problems like yours with this approach:

first sort it as:

$f(x,y) = 0$

then use this formula to differntiate:

$\frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}$

it means when you are differntiatig a respect to x then y assumed as constant and in reverse; be sure before differniation that all statements of the equation should be sored on one side and other side should be only Zero(0).

**11rdc11** · October 25th 2008, 11:26 PM

Originally Posted by toraj58

as a genral solution you can sovle all problems like yours with this approach:

first sort it as:

$f(x,y) = 0$

then use this formula to differntiate:

$\frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}$

it means when you are differntiatig a respect to x then y assumed as constant and in reverse; be sure before differniation that all statements of the equation should be sored on one side and other side should be only Zero(0).

I never understood why we were taught that so late in the course. I find it easier than doing implicit diff

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