Asked by Anonymous
y=tan(ln(2x+1))
Derive
Derive
Answers
Answered by
Motorhead
Let u = 2x+1
Let v = ln u
Then y = tan v
dy/dx = (dy/dv)(dv/du)(du/dx)
dy/dx = (sec²v)(1/u)(2)
dy/dx = (sec²(ln (2x+1))(1/(2x+1))(2)
dy/dx = 2sec²(ln (2x+1))/(2x+1)
Let v = ln u
Then y = tan v
dy/dx = (dy/dv)(dv/du)(du/dx)
dy/dx = (sec²v)(1/u)(2)
dy/dx = (sec²(ln (2x+1))(1/(2x+1))(2)
dy/dx = 2sec²(ln (2x+1))/(2x+1)
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