Find the derivative of ln & e

(1)
y = x (ln x)^2

dy/dx
= d[x (ln x)^2]
= (x)d[(ln x)^2] + [(ln x)^2]d(x)
= x[2(In x)]d(In x) + (In x)^2
= 2x(In x)(1/x) + (In x)^2
= 2Inx + (In x)^2
= (In x)(2 + In x)



(2)
y = (e^x - e^-x) / (e^x+e^-x)


=> y = { [e^(-x)] [e^(2x) -1] } / { [e^(-x)] [e^(2x)+1 ] }
=> y = [e^(2x)-1] / [e^(2x)+1]


Consider
In y = In[e^(2x)-1] - In[e^(2x)+1]

Differentiate both sides,
=> (1/y)(dy/dx) = {1/[e^(2x)-1] }d[e^(2x)-1]- {1/[e^(2x)+1] } d[e^(2x)+1]
=> (1/y)(dy/dx) = [2e^(2x)] / [e^(2x)-1] - [2e^2x] / [e^(2x)+1]
=> dy/dx = (y)[2e^(2x)] {1/ [e^(2x)-1] - 1/[e^(2x)+1]}
=> dy/dx = {[e^(2x)-1] / [e^(2x)+1]} [2e^(2x)] {1/ [e^(2x)-1] - 1/[e^(2x)+1]}
=> dy/dx = {[e^(2x)-1][2e^(2x)] }{1/ [e^(2x)-1] - 1/[e^(2x)+1]} / [e^(2x)+1]
=> dy/dx = {[e^(2x)-1][2e^(2x)]/ [e^(2x)+1]}{[e^(2x)+1] - [e^(2x)-1]} / {[e^(2x)+1][e^(2x)-1]}
=> dy/dx = {[e^(2x)-1][2e^(2x)]/ [e^(2x)+1]}{ 2 / {[e^(2x)+1][e^(2x)-1]} }
=> dy/dx = {2[e^(2x)-1][2e^(2x)]} / {[e^(2x)+1][e^(2x)+1][e^(2x)-1]}
=> dy/dx = {2[2e^(2x)]} / {[e^(2x)+1][e^(2x)+1]}
=> dy/dx = 4e^(2x) / { [e^(2x)+1]^2 }

(1) y = x (ln x)2
dy/dx = (ln x)2 + x d[(ln x)2]/dx
= (ln x)2 + 2x ln x d(ln x)dx
= (ln x)2 + 2 lnx
(2) y = (ex - e-x)/(ex + e-x)
ln y = ln (ex - e-x) - ln (ex + e-x)
y'/y = [1/(ex - e-x)] d(ex - e-x)/dx - [1/(ex + e-x)] d(ex + e-x)/dx
= (ex + e-x)/(ex - e-x) - (ex - e-x)/(ex + e-x)
= [(ex + e-x)2 - (ex - e-x)2]/[(ex - e-x)(ex + e-x)]
= 4/[(ex - e-x)(ex + e-x)]
y' = 4y/[(ex - e-x)(ex + e-x)]
dy/dx = 4/(ex + e-x)2

我英文程度不好，所以用中文作答，請見諒。
1.
這有用到兩個觀念
(1) [f(x)g(x)]'=f(x)'g(x)+f(x)g(x)'
(2) [f(g(x))]'=f'(g(x))×g(x)'
∴dy=[(lnx)^2+x×2(lnx)×(1/x)]dx
dy=[(lnx)^2+2×(lnx)] dx

2.
sinhx=(e^x - e^-x)/2
coshx=(e^x+e^-x)/2
∴原題目可以改寫成：
y=sinhx/coshx
dy=[(coshx)^2-(sinhx)^2]/(coshx)^2
因為(coshx)^2-(sinhx)^2=1、1/coshx=sechx
∴dy=(sechx)^2 dx

如果有不懂歡迎再詢問喔~^^

(1) y = x (ln x)2
dy/dx = (ln x)2 + x d[(ln x)2]/dx
= (ln x)2 + 2x ln x d(ln x)dx
= (ln x)2 + 2 lnx
(2) y = (ex - e-x)/(ex + e-x)
ln y = ln (ex - e-x) - ln (ex + e-x)
y'/y = [1/(ex - e-x)] d(ex - e-x)/dx - [1/(ex + e-x)] d(ex + e-x)/dx
= (ex + e-x)/(ex - e-x) - (ex - e-x)/(ex + e-x)
= [(ex + e-x)2 - (ex - e-x)2]/[(ex - e-x)(ex + e-x)]
= 4/[(ex - e-x)(ex + e-x)]
y' = 4y/[(ex - e-x)(ex + e-x)]
dy/dx = 4/(ex + e-x)2