Differentiate #y=cosh^(-1)(sinhx)#?

Question

5041 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-01T23:03:24

#y=cosh^(-1) (sinhx)#

#coshy=sinhx#

#y'*sinhy=coshx#

#y'=coshx/sinhy#

#y'=coshx/sqrt[(coshy)^2-1]#

#y'=coshx/sqrt[(sinhx)^2-1]#

1) I transformed #y=cosh^(-1) (sinhx)# into #coshy=sinhx#.

2) I took differentiation both sides.

3) I left #y'# alone dividing both sides by #sinhy#.

4) I replaced #sinhy# with hyperbolic function in terms of x using #sinhy=sqrt[(coshy)^2-1]=sqrt[(sinhx)^2-1]# equation.