Finding the derivative of F(t)=ln(arctan(t)). Can someone please help me find the derivative of this function? Thank you in advance!?
回答 (4)
2016-07-19 4:54 am
✔ 最佳答案
F(t) = ln(tan⁻¹(t))
F'(t) = 1/tan⁻¹(t) * d/dt (tan⁻¹(t))
F'(t) = 1/tan⁻¹(t) * 1/(t²+1)
F'(t) = 1 / [(t²+1) tan⁻¹(t)]
2016-07-19 4:58 am
F(t) = ln[arctan(t)]
F'(t) = d{ln[arctan(t)]}/dt
= d{ln[arctan(t)]}/d[arctan(t)] * d[arctan(t)]/dt
= [1/arctan(t)] * [1/(1+t²)]
= 1/[(1+t²) arctan(t)]
F'(t) = d{ln[arctan(t)]}/dt
= d{ln[arctan(t)]}/d[arctan(t)] * d[arctan(t)]/dt
= [1/arctan(t)] * [1/(1+t²)]
= 1/[(1+t²) arctan(t)]
2016-07-19 7:14 am
1/arctan(t)(t^2+1)
see step by step solution:
https://www.symbolab.com/solver/derivative-calculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(ln%5Cleft(arctan%5Cleft(t%5Cright)%5Cright)%5Cright)
hope this helps
see step by step solution:
https://www.symbolab.com/solver/derivative-calculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(ln%5Cleft(arctan%5Cleft(t%5Cright)%5Cright)%5Cright)
hope this helps
2016-07-19 4:54 am
y = ln(arctan(t))
e^y = arctan(t)
tan(e^y) = t
Derive implicitly and solve for dy/dt
e^(y) * sec(e^y)^2 * dy = dt
dy/dt = 1 / (e^(y) * sec(e^y)^2)
dy/dt = 1 / (e^(y) * (1 + tan(e^y)^2))
dy/dt = 1 / (arctan(t) * (1 + t^2))
e^y = arctan(t)
tan(e^y) = t
Derive implicitly and solve for dy/dt
e^(y) * sec(e^y)^2 * dy = dt
dy/dt = 1 / (e^(y) * sec(e^y)^2)
dy/dt = 1 / (e^(y) * (1 + tan(e^y)^2))
dy/dt = 1 / (arctan(t) * (1 + t^2))
收錄日期: 2021-04-18 15:20:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160718205046AAgqZut