using the derivative of e^x,show that if y=ln(x),then y'=1/x?
回答 (4)
2018-12-06 11:38 pm
e^y = x
e^y dy/dx = 1
dy/dx = 1 / e^y
dy/dx = 1 / x
e^y dy/dx = 1
dy/dx = 1 / e^y
dy/dx = 1 / x
2018-12-06 11:36 pm
y = ln(x)
e^y = e^ln(x)
e^y = x
Derive implicitly
e^(y) * dy = dx
Solve for dy/dx
dy/dx = 1 / e^(y)
We know that e^(y) = x, so
dy/dx = 1/x
e^y = e^ln(x)
e^y = x
Derive implicitly
e^(y) * dy = dx
Solve for dy/dx
dy/dx = 1 / e^(y)
We know that e^(y) = x, so
dy/dx = 1/x
2018-12-07 1:03 am
If y = ln(x) then by definition
e^y = x, then differentiate wrt y
e^(y) = dx/dy = x
dy/dx = 1/x
e^y = x, then differentiate wrt y
e^(y) = dx/dy = x
dy/dx = 1/x
2018-12-07 12:24 am
y=ln(x)
=>
e^y=x
=>
(e^y)y'=1
=>
y'=1/e^y
=>
y'=1/e^(ln(x)
=>
y'=1/x
=>
e^y=x
=>
(e^y)y'=1
=>
y'=1/e^y
=>
y'=1/e^(ln(x)
=>
y'=1/x
收錄日期: 2021-04-24 01:13:50
原文連結 [永久失效]:
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