Asked by Hailey
find y'(x) if x^y=y^x
Answers
Answered by
Count Iblis
x^y=y^x --->
y log(x) = x log(y) ----->
log(x) dy + y/x dx =
log(y) dx + x/y dy ---->
[log(x) - x/y] dy = [log(y) - y/x] dx
---->
dy/dx = [log(y) - y/x]/[log(x) - x/y]
y log(x) = x log(y) ----->
log(x) dy + y/x dx =
log(y) dx + x/y dy ---->
[log(x) - x/y] dy = [log(y) - y/x] dx
---->
dy/dx = [log(y) - y/x]/[log(x) - x/y]
Answered by
Bosnian
Go on :
wolframalpha dot com
When page be open in rectangle type:
derivative x^y=y^x
and click option =
When you see result click option:
Show steps
wolframalpha dot com
When page be open in rectangle type:
derivative x^y=y^x
and click option =
When you see result click option:
Show steps
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