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Having trouble solving implicit function

0

$begingroup$

$$ xcos y + ycos u + ucos x = 1 $$

I'm trying to calculate u$'_x$ and $u'_y$ in the point $(0,1)$
while $0 leq u leq pi$.

I've managed to find:

$$u'_x = - fraccos y-usin xcos x-ysin u\
u'_y = - fraccos u-xsin ycos x-ysin u$$

What holds me back currently I can't find $u$.

I get to this equation when I place $(0,1)$ ----> $ cos u + u = 1$

Thanks in advance, hope this question isn't too dumb.

calculus derivatives implicit-differentiation

share|cite|improve this question

edited 2 days ago

Brian

1,202116

asked 2 days ago

ShamesShames

83

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Drawing the graphs of $umapstocos u$ and $umapsto 1-u$ you can see that necessarily $u=0$ at the point in question.
  $endgroup$
  – Christian Blatter
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  note: $cos u<1-u, 0<ule fracpi2$ and $cos 0=1-0$. See the desmos graph.
  $endgroup$
  – farruhota
  2 days ago
  

  
  
  
  

add a comment | 

0

$begingroup$

$$ xcos y + ycos u + ucos x = 1 $$

I'm trying to calculate u$'_x$ and $u'_y$ in the point $(0,1)$
while $0 leq u leq pi$.

I've managed to find:

$$u'_x = - fraccos y-usin xcos x-ysin u\
u'_y = - fraccos u-xsin ycos x-ysin u$$

What holds me back currently I can't find $u$.

I get to this equation when I place $(0,1)$ ----> $ cos u + u = 1$

Thanks in advance, hope this question isn't too dumb.

calculus derivatives implicit-differentiation

share|cite|improve this question

edited 2 days ago

Brian

1,202116

asked 2 days ago

ShamesShames

83

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Drawing the graphs of $umapstocos u$ and $umapsto 1-u$ you can see that necessarily $u=0$ at the point in question.
  $endgroup$
  – Christian Blatter
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  note: $cos u<1-u, 0<ule fracpi2$ and $cos 0=1-0$. See the desmos graph.
  $endgroup$
  – farruhota
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

$$ xcos y + ycos u + ucos x = 1 $$

I'm trying to calculate u$'_x$ and $u'_y$ in the point $(0,1)$
while $0 leq u leq pi$.

I've managed to find:

$$u'_x = - fraccos y-usin xcos x-ysin u\
u'_y = - fraccos u-xsin ycos x-ysin u$$

What holds me back currently I can't find $u$.

I get to this equation when I place $(0,1)$ ----> $ cos u + u = 1$

Thanks in advance, hope this question isn't too dumb.

calculus derivatives implicit-differentiation

share|cite|improve this question

edited 2 days ago

Brian

1,202116

asked 2 days ago

ShamesShames

83

$endgroup$

share|cite|improve this question

edited 2 days ago

Brian

1,202116

asked 2 days ago

ShamesShames

83

share|cite|improve this question

edited 2 days ago

Brian

1,202116

asked 2 days ago

ShamesShames

83

edited 2 days ago

Brian

1,202116

edited 2 days ago

Brian

1,202116

asked 2 days ago

ShamesShames

83

asked 2 days ago

ShamesShames

83