Asked by Matt
how can I calculate this deriivative:
1/(sin(x-sinx))?
1/(sin(x-sinx))?
Answers
Answered by
Count Iblis
Chain rule:
d/dx [1/(sin(x-sinx))] =
-1/[sin^2(x-sin(x)] *
d/dx [sin(x-sin(x))] =
-cos(x-sin(x))/[sin^2(x-sin(x)]*
d/dx [x-sin(x)] =
[cos(x)-1]cos(x-sin(x))/[sin^2(x-sin(x)]
d/dx [1/(sin(x-sinx))] =
-1/[sin^2(x-sin(x)] *
d/dx [sin(x-sin(x))] =
-cos(x-sin(x))/[sin^2(x-sin(x)]*
d/dx [x-sin(x)] =
[cos(x)-1]cos(x-sin(x))/[sin^2(x-sin(x)]
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