# Math Help - differentiation help..

1. ## differentiation help..

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

2. 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.

3. 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?

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

5. 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

6. ## help

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

7. 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?

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

8. 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

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

10. 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...

11. 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

12. 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

13. 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

14. 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).

15. 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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