How do you find the derivative of x = cos(y)? [closed]

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By just deriving, I would get $-csc(y)$ but apparently it's $y=arccos(x)$? Can someone explain why and how they did it?

edited Sep 5 at 11:36

bjcolby15

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asked Sep 5 at 6:34

okskkkk

closed as off-topic by user91500, Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh Sep 5 at 13:07

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up vote
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down vote

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By just deriving, I would get $-csc(y)$ but apparently it's $y=arccos(x)$? Can someone explain why and how they did it?

edited Sep 5 at 11:36

bjcolby15

8671816

asked Sep 5 at 6:34

okskkkk

closed as off-topic by user91500, Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh Sep 5 at 13:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

add a commentÂ |Â

up vote
-4
down vote

favorite

By just deriving, I would get $-csc(y)$ but apparently it's $y=arccos(x)$? Can someone explain why and how they did it?

edited Sep 5 at 11:36

bjcolby15

8671816

asked Sep 5 at 6:34

okskkkk

By just deriving, I would get $-csc(y)$ but apparently it's $y=arccos(x)$? Can someone explain why and how they did it?

calculus trigonometry implicit-differentiation

edited Sep 5 at 11:36

bjcolby15

8671816

asked Sep 5 at 6:34

okskkkk

edited Sep 5 at 11:36

bjcolby15

8671816

asked Sep 5 at 6:34

okskkkk

edited Sep 5 at 11:36

bjcolby15

8671816

edited Sep 5 at 11:36

bjcolby15

8671816

edited Sep 5 at 11:36

bjcolby15

8671816

asked Sep 5 at 6:34

okskkkk

asked Sep 5 at 6:34

okskkkk

asked Sep 5 at 6:34

okskkkk

closed as off-topic by user91500, Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh Sep 5 at 13:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by user91500, Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh Sep 5 at 13:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Delta-u, amWhy, Jose Arnaldo Bebita Dris, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

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1 Answer
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$x=cos(y)$ gives me $ fracdxdy = -sin(y)$. Shuffeling around gives then $ fracdydx = - csc(y)$. But we know that $x=cos(y)$ so we get $y = arccos(x)$. This gives then $fracdydx = -csc(arccos(x))$.

And hint for in the future: Show us what work you did and try to formulate your question is a better way!

edited Sep 5 at 8:20

Babelfish

1,004115

answered Sep 5 at 7:08

dani

2229

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

$x=cos(y)$ gives me $ fracdxdy = -sin(y)$. Shuffeling around gives then $ fracdydx = - csc(y)$. But we know that $x=cos(y)$ so we get $y = arccos(x)$. This gives then $fracdydx = -csc(arccos(x))$.

And hint for in the future: Show us what work you did and try to formulate your question is a better way!

edited Sep 5 at 8:20

Babelfish

1,004115

answered Sep 5 at 7:08

dani

2229

add a commentÂ |Â

up vote
1
down vote

$x=cos(y)$ gives me $ fracdxdy = -sin(y)$. Shuffeling around gives then $ fracdydx = - csc(y)$. But we know that $x=cos(y)$ so we get $y = arccos(x)$. This gives then $fracdydx = -csc(arccos(x))$.

And hint for in the future: Show us what work you did and try to formulate your question is a better way!

edited Sep 5 at 8:20

Babelfish

1,004115

answered Sep 5 at 7:08

dani

2229

add a commentÂ |Â

up vote
1
down vote

$x=cos(y)$ gives me $ fracdxdy = -sin(y)$. Shuffeling around gives then $ fracdydx = - csc(y)$. But we know that $x=cos(y)$ so we get $y = arccos(x)$. This gives then $fracdydx = -csc(arccos(x))$.

And hint for in the future: Show us what work you did and try to formulate your question is a better way!

edited Sep 5 at 8:20

Babelfish

1,004115

answered Sep 5 at 7:08

dani

2229

$x=cos(y)$ gives me $ fracdxdy = -sin(y)$. Shuffeling around gives then $ fracdydx = - csc(y)$. But we know that $x=cos(y)$ so we get $y = arccos(x)$. This gives then $fracdydx = -csc(arccos(x))$.

And hint for in the future: Show us what work you did and try to formulate your question is a better way!

edited Sep 5 at 8:20

Babelfish

1,004115

answered Sep 5 at 7:08

dani

2229

edited Sep 5 at 8:20

Babelfish

1,004115

edited Sep 5 at 8:20

Babelfish

1,004115

edited Sep 5 at 8:20

Babelfish

1,004115

answered Sep 5 at 7:08

dani

2229

answered Sep 5 at 7:08

dani

2229

answered Sep 5 at 7:08

dani

2229

add a commentÂ |Â

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