plz help me differentiate

Question

arccosh(x/√3)
i no that from inverse heyperbolic function
cosh=(x±√(x²-1)
but i still dont know how to that go about it

Answer 1

just use the chain rule

d/dx arccosh(u) = 1/√(u^2-1) du/dx
so, if u = x/√3, that becomes

1/√(x^2/3 - 1) * 1/√3
= 1 / √(x^2-3)

since you know that

cosh^2(u) = 1+sinh^2(u)
d/dx coshu = sinhu, you have
y = arccosh(u)
coshy = u
sinhy y' = 1
y' = 1/sinhy = 1/√(u^2-1)

Now, if you want to do it the hard way, from

y = arccosh(u) = ln(u + √(u^2-1))

then taking derivatives,

y' = 1/(u+√(u^2-1)) (1 + u/√(u^2-1))
= (√(u^2-1) + u) / (u+√(u^2-1))√(u^2-1)
= 1/√(u^2-1)

Answer 2

blesssssssssssssssssssss u sir thank thank thank u