plz help me differentiate

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

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)

blesssssssssssssssssssss u sir thank thank thank u

To differentiate the function arccosh(x/√3), we can use the chain rule of differentiation.

First, let's simplify the expression inside the inverse hyperbolic function.

We have arccosh(x/√3).

Now, recall the identity cosh^2(x) - sinh^2(x) = 1 for hyperbolic functions.

In our case, we can rewrite the expression as cosh^2(arccosh(x/√3)) - sinh^2(arccosh(x/√3)) = 1.

Since cosh(arccosh(x/√3)) is equal to (x/√3) and sinh(arccosh(x/√3)) is the square root of (cosh^2(arccosh(x/√3)) - 1), we have:

(x^2/3) - sinh^2(arccosh(x/√3)) = 1.

Now, we can differentiate both sides of the equation.

On the left side, differentiating (x^2/3) with respect to x gives us (2x/3).

On the right side, differentiating 1 gives us 0.

Differentiating sinh^2(arccosh(x/√3)) requires the chain rule. Let u = arccosh(x/√3).

We then have the derivative of sinh^2(u):

d/dx (sinh^2(u)) = d/dx (sinh(u))^2 = 2sinh(u) * d/dx (sinh(u)) = 2sinh(u) * cosh(u) * du/dx.

Substituting u = arccosh(x/√3) into the equation above, we get:

d/dx (sinh^2(arccosh(x/√3))) = 2sinh(arccosh(x/√3)) * cosh(arccosh(x/√3)) * du/dx.

Now, we can substitute back the values we know:

d/dx (sinh^2(arccosh(x/√3))) = 2sinh(arccosh(x/√3)) * cosh(arccosh(x/√3)) * (d/dx (arccosh(x/√3))).

Finally, we can substitute back the original expression and solve for the derivative:

d/dx (arccosh(x/√3)) = (2x/√(3^2 - x^2)).